Aljabar Contoh

$f (x) = 10 x^{\frac{1}{5}} - 2$

Langkah 1

Tuliskan $f (x) = 10 x^{\frac{1}{5}} - 2$ sebagai sebuah persamaan.

$y = 10 x^{\frac{1}{5}} - 2$

Langkah 2

Saling tukar variabel.

$x = 10 y^{\frac{1}{5}} - 2$

Langkah 3

Selesaikan $y$ .

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Langkah 3.1

Tulis kembali persamaan tersebut sebagai $10 y^{\frac{1}{5}} - 2 = x$ .

$10 y^{\frac{1}{5}} - 2 = x$

Langkah 3.2

Pindahkan semua suku yang tidak mengandung $y$ ke sisi kanan dari persamaan.

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Langkah 3.2.1

Tambahkan $2$ ke kedua sisi persamaan.

$10 y^{\frac{1}{5}} - x = 2$

Langkah 3.2.2

Tambahkan $x$ ke kedua sisi persamaan.

$10 y^{\frac{1}{5}} = 2 + x$

Langkah 3.3

Pangkatkan setiap sisi persamaan dengan pangkat $5$ untuk menghilangkan eksponen pecahan di sisi kiri.

${(10 y^{\frac{1}{5}})}^{5} = {(2 + x)}^{5}$

Langkah 3.4

Sederhanakan bentuk eksponen.

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Langkah 3.4.1

Sederhanakan sisi kirinya.

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Langkah 3.4.1.1

Sederhanakan ${(10 y^{\frac{1}{5}})}^{5}$ .

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Langkah 3.4.1.1.1

Terapkan kaidah hasil kali ke $10 y^{\frac{1}{5}}$ .

$10^{5} {(y^{\frac{1}{5}})}^{5} = {(2 + x)}^{5}$

Langkah 3.4.1.1.2

Naikkan $10$ menjadi pangkat $5$ .

$100000 {(y^{\frac{1}{5}})}^{5} = {(2 + x)}^{5}$

Langkah 3.4.1.1.3

Kalikan eksponen dalam ${(y^{\frac{1}{5}})}^{5}$ .

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Langkah 3.4.1.1.3.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$100000 y^{\frac{1}{5} \cdot 5} = {(2 + x)}^{5}$

Langkah 3.4.1.1.3.2

Batalkan faktor persekutuan dari $5$ .

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Langkah 3.4.1.1.3.2.1

Batalkan faktor persekutuan.

$100000 y^{\frac{1}{5} \cdot 5} = {(2 + x)}^{5}$

Langkah 3.4.1.1.3.2.2

Tulis kembali pernyataannya.

$100000 y^{1} = {(2 + x)}^{5}$

Langkah 3.4.1.1.4

Sederhanakan.

$100000 y = {(2 + x)}^{5}$

Langkah 3.4.2

Sederhanakan sisi kanannya.

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Langkah 3.4.2.1

Sederhanakan ${(2 + x)}^{5}$ .

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Langkah 3.4.2.1.1

Gunakan Teorema Binomial.

$100000 y = 2^{5} + 5 \cdot 2^{4} x + 10 \cdot 2^{3} x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2

Sederhanakan setiap suku.

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Langkah 3.4.2.1.2.1

Naikkan $2$ menjadi pangkat $5$ .

$100000 y = 32 + 5 \cdot 2^{4} x + 10 \cdot 2^{3} x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.2

Naikkan $2$ menjadi pangkat $4$ .

$100000 y = 32 + 5 \cdot 16 x + 10 \cdot 2^{3} x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.3

Kalikan $5$ dengan $16$ .

$100000 y = 32 + 80 x + 10 \cdot 2^{3} x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.4

Naikkan $2$ menjadi pangkat $3$ .

$100000 y = 32 + 80 x + 10 \cdot 8 x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.5

Kalikan $10$ dengan $8$ .

$100000 y = 32 + 80 x + 80 x^{2} + 10 \cdot 2^{2} x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.6

Naikkan $2$ menjadi pangkat $2$ .

$100000 y = 32 + 80 x + 80 x^{2} + 10 \cdot 4 x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.7

Kalikan $10$ dengan $4$ .

$100000 y = 32 + 80 x + 80 x^{2} + 40 x^{3} + 5 \cdot 2 x^{4} + x^{5}$

Langkah 3.4.2.1.2.8

Kalikan $5$ dengan $2$ .

$100000 y = 32 + 80 x + 80 x^{2} + 40 x^{3} + 10 x^{4} + x^{5}$

Langkah 3.5

Bagi setiap suku pada $100000 y = 32 + 80 x + 80 x^{2} + 40 x^{3} + 10 x^{4} + x^{5}$ dengan $100000$ dan sederhanakan.

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Langkah 3.5.1

Bagilah setiap suku di $100000 y = 32 + 80 x + 80 x^{2} + 40 x^{3} + 10 x^{4} + x^{5}$ dengan $100000$ .

$\frac{100000 y}{100000} = \frac{32}{100000} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.2

Sederhanakan sisi kirinya.

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Langkah 3.5.2.1

Batalkan faktor persekutuan dari $100000$ .

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Langkah 3.5.2.1.1

Batalkan faktor persekutuan.

$\frac{100000 y}{100000} = \frac{32}{100000} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.2.1.2

Bagilah $y$ dengan $1$ .

$y = \frac{32}{100000} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3

Sederhanakan sisi kanannya.

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Langkah 3.5.3.1

Sederhanakan setiap suku.

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Langkah 3.5.3.1.1

Hapus faktor persekutuan dari $32$ dan $100000$ .

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Langkah 3.5.3.1.1.1

Faktorkan $32$ dari $32$ .

$y = \frac{32 (1)}{100000} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.1.2

Batalkan faktor persekutuan.

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Langkah 3.5.3.1.1.2.1

Faktorkan $32$ dari $100000$ .

$y = \frac{32 \cdot 1}{32 \cdot 3125} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.1.2.2

Batalkan faktor persekutuan.

$y = \frac{32 \cdot 1}{32 \cdot 3125} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.1.2.3

Tulis kembali pernyataannya.

$y = \frac{1}{3125} + \frac{80 x}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.2

Hapus faktor persekutuan dari $80$ dan $100000$ .

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Langkah 3.5.3.1.2.1

Faktorkan $80$ dari $80 x$ .

$y = \frac{1}{3125} + \frac{80 (x)}{100000} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.2.2

Batalkan faktor persekutuan.

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Langkah 3.5.3.1.2.2.1

Faktorkan $80$ dari $100000$ .

$y = \frac{1}{3125} + \frac{80 x}{80 \cdot 1250} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.2.2.2

Batalkan faktor persekutuan.

$y = \frac{1}{3125} + \frac{80 x}{80 \cdot 1250} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.2.2.3

Tulis kembali pernyataannya.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{80 x^{2}}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.3

Hapus faktor persekutuan dari $80$ dan $100000$ .

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Langkah 3.5.3.1.3.1

Faktorkan $80$ dari $80 x^{2}$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{80 (x^{2})}{100000} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.3.2

Batalkan faktor persekutuan.

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Langkah 3.5.3.1.3.2.1

Faktorkan $80$ dari $100000$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{80 x^{2}}{80 \cdot 1250} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.3.2.2

Batalkan faktor persekutuan.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{80 x^{2}}{80 \cdot 1250} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.3.2.3

Tulis kembali pernyataannya.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{40 x^{3}}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.4

Hapus faktor persekutuan dari $40$ dan $100000$ .

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Langkah 3.5.3.1.4.1

Faktorkan $40$ dari $40 x^{3}$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{40 (x^{3})}{100000} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.4.2

Batalkan faktor persekutuan.

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Langkah 3.5.3.1.4.2.1

Faktorkan $40$ dari $100000$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{40 x^{3}}{40 \cdot 2500} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.4.2.2

Batalkan faktor persekutuan.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{40 x^{3}}{40 \cdot 2500} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.4.2.3

Tulis kembali pernyataannya.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{10 x^{4}}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.5

Hapus faktor persekutuan dari $10$ dan $100000$ .

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Langkah 3.5.3.1.5.1

Faktorkan $10$ dari $10 x^{4}$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{10 (x^{4})}{100000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.5.2

Batalkan faktor persekutuan.

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Langkah 3.5.3.1.5.2.1

Faktorkan $10$ dari $100000$ .

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{10 x^{4}}{10 \cdot 10000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.5.2.2

Batalkan faktor persekutuan.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{10 x^{4}}{10 \cdot 10000} + \frac{x^{5}}{100000}$

Langkah 3.5.3.1.5.2.3

Tulis kembali pernyataannya.

$y = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}$

Langkah 4

Ganti $y$ dengan $f^{- 1} (x)$ untuk memunculkan jawaban akhir.

$f^{- 1} (x) = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}$

Langkah 5

Periksa apakah $f^{- 1} (x) = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}$ merupakan balikan dari $f (x) = 10 x^{\frac{1}{5}} - 2$ .

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Langkah 5.1

Untuk memverifikasi balikannya, periksa apakah $f^{- 1} (f (x)) = x$ dan $f (f^{- 1} (x)) = x$ .

Langkah 5.2

Evaluasi $f^{- 1} (f (x))$ .

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Langkah 5.2.1

Tulis fungsi hasil komposit.

$f^{- 1} (f (x))$

Langkah 5.2.2

Evaluasi $f^{- 1} (10 x^{\frac{1}{5}} - 2)$ dengan mensubstitusikan nilai (Variabel1) ke dalam (Variabel2).

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{10 x^{\frac{1}{5}} - 2}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3

Sederhanakan suku-suku.

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Langkah 5.2.3.1

Sederhanakan setiap suku.

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Langkah 5.2.3.1.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{2 (5 x^{\frac{1}{5}}) - 2}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.2

Faktorkan $2$ dari $- 2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{2 (5 x^{\frac{1}{5}}) + 2 (- 1)}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.3

Faktorkan $2$ dari $2 (5 x^{\frac{1}{5}}) + 2 (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{2 (5 x^{\frac{1}{5}} - 1)}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.4

Batalkan faktor persekutuan.

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Langkah 5.2.3.1.4.1

Faktorkan $2$ dari $1250$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{2 (5 x^{\frac{1}{5}} - 1)}{2 (625)} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.4.2

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{2 (5 x^{\frac{1}{5}} - 1)}{2 \cdot 625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.4.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.5

Sederhanakan pembilangnya.

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Langkah 5.2.3.1.5.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}} - 2$ .

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Langkah 5.2.3.1.5.1.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) - 2)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.5.1.2

Faktorkan $2$ dari $- 2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) + 2 (- 1))}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.5.1.3

Faktorkan $2$ dari $2 (5 x^{\frac{1}{5}}) + 2 (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{{(2 (5 x^{\frac{1}{5}} - 1))}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.5.2

Terapkan kaidah hasil kali ke $2 (5 x^{\frac{1}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2^{2} {(5 x^{\frac{1}{5}} - 1)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.5.3

Naikkan $2$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{4 {(5 x^{\frac{1}{5}} - 1)}^{2}}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.6

Faktorkan $2$ dari $4 {(5 x^{\frac{1}{5}} - 1)}^{2}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 (2 {(5 x^{\frac{1}{5}} - 1)}^{2})}{1250} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.7

Batalkan faktor persekutuan.

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Langkah 5.2.3.1.7.1

Faktorkan $2$ dari $1250$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 (2 {(5 x^{\frac{1}{5}} - 1)}^{2})}{2 (625)} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.7.2

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 (2 {(5 x^{\frac{1}{5}} - 1)}^{2})}{2 \cdot 625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.7.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.8

Sederhanakan pembilangnya.

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Langkah 5.2.3.1.8.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}} - 2$ .

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Langkah 5.2.3.1.8.1.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) - 2)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.8.1.2

Faktorkan $2$ dari $- 2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) + 2 (- 1))}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.8.1.3

Faktorkan $2$ dari $2 (5 x^{\frac{1}{5}}) + 2 (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{{(2 (5 x^{\frac{1}{5}} - 1))}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.8.2

Terapkan kaidah hasil kali ke $2 (5 x^{\frac{1}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2^{3} {(5 x^{\frac{1}{5}} - 1)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.8.3

Naikkan $2$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{8 {(5 x^{\frac{1}{5}} - 1)}^{3}}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.9

Faktorkan $4$ dari $8 {(5 x^{\frac{1}{5}} - 1)}^{3}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{4 (2 {(5 x^{\frac{1}{5}} - 1)}^{3})}{2500} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.10

Batalkan faktor persekutuan.

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Langkah 5.2.3.1.10.1

Faktorkan $4$ dari $2500$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{4 (2 {(5 x^{\frac{1}{5}} - 1)}^{3})}{4 (625)} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.10.2

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{4 (2 {(5 x^{\frac{1}{5}} - 1)}^{3})}{4 \cdot 625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.10.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.11

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Langkah 5.2.3.1.11.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}} - 2$ .

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Langkah 5.2.3.1.11.1.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) - 2)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.11.1.2

Faktorkan $2$ dari $- 2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) + 2 (- 1))}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.11.1.3

Faktorkan $2$ dari $2 (5 x^{\frac{1}{5}}) + 2 (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(2 (5 x^{\frac{1}{5}} - 1))}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.11.2

Terapkan kaidah hasil kali ke $2 (5 x^{\frac{1}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{2^{4} {(5 x^{\frac{1}{5}} - 1)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.11.3

Naikkan $2$ menjadi pangkat $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{16 {(5 x^{\frac{1}{5}} - 1)}^{4}}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.12

Faktorkan $16$ dari $16 {(5 x^{\frac{1}{5}} - 1)}^{4}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{16 ({(5 x^{\frac{1}{5}} - 1)}^{4})}{10000} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.13

Batalkan faktor persekutuan.

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Langkah 5.2.3.1.13.1

Faktorkan $16$ dari $10000$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{16 {(5 x^{\frac{1}{5}} - 1)}^{4}}{16 \cdot 625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.13.2

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{16 {(5 x^{\frac{1}{5}} - 1)}^{4}}{16 \cdot 625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.13.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{{(10 x^{\frac{1}{5}} - 2)}^{5}}{100000}$

Langkah 5.2.3.1.14

Sederhanakan pembilangnya.

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Langkah 5.2.3.1.14.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}} - 2$ .

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Langkah 5.2.3.1.14.1.1

Faktorkan $2$ dari $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) - 2)}^{5}}{100000}$

Langkah 5.2.3.1.14.1.2

Faktorkan $2$ dari $- 2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{{(2 (5 x^{\frac{1}{5}}) + 2 (- 1))}^{5}}{100000}$

Langkah 5.2.3.1.14.1.3

Faktorkan $2$ dari $2 (5 x^{\frac{1}{5}}) + 2 (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{{(2 (5 x^{\frac{1}{5}} - 1))}^{5}}{100000}$

Langkah 5.2.3.1.14.2

Terapkan kaidah hasil kali ke $2 (5 x^{\frac{1}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{2^{5} {(5 x^{\frac{1}{5}} - 1)}^{5}}{100000}$

Langkah 5.2.3.1.14.3

Naikkan $2$ menjadi pangkat $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{32 {(5 x^{\frac{1}{5}} - 1)}^{5}}{100000}$

Langkah 5.2.3.1.15

Faktorkan $32$ dari $32 {(5 x^{\frac{1}{5}} - 1)}^{5}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{32 ({(5 x^{\frac{1}{5}} - 1)}^{5})}{100000}$

Langkah 5.2.3.1.16

Batalkan faktor persekutuan.

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Langkah 5.2.3.1.16.1

Faktorkan $32$ dari $100000$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{32 {(5 x^{\frac{1}{5}} - 1)}^{5}}{32 \cdot 3125}$

Langkah 5.2.3.1.16.2

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{32 {(5 x^{\frac{1}{5}} - 1)}^{5}}{32 \cdot 3125}$

Langkah 5.2.3.1.16.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{5 x^{\frac{1}{5}} - 1}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{2}}{625} + \frac{2 {(5 x^{\frac{1}{5}} - 1)}^{3}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{4}}{625} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125}$

Langkah 5.2.3.2

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 {(5 x^{\frac{1}{5}} - 1)}^{2} + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3

Sederhanakan setiap suku.

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Langkah 5.2.3.3.1

Tulis kembali ${(5 x^{\frac{1}{5}} - 1)}^{2}$ sebagai $(5 x^{\frac{1}{5}} - 1) (5 x^{\frac{1}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 ((5 x^{\frac{1}{5}} - 1) (5 x^{\frac{1}{5}} - 1)) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.2

Perluas $(5 x^{\frac{1}{5}} - 1) (5 x^{\frac{1}{5}} - 1)$ menggunakan Metode FOIL.

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Langkah 5.2.3.3.2.1

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 x^{\frac{1}{5}} (5 x^{\frac{1}{5}} - 1) - 1 (5 x^{\frac{1}{5}} - 1)) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.2.2

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 x^{\frac{1}{5}} (5 x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}} - 1)) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.2.3

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 x^{\frac{1}{5}} (5 x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3

Sederhanakan dan gabungkan suku-suku sejenis.

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Langkah 5.2.3.3.3.1

Sederhanakan setiap suku.

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Langkah 5.2.3.3.3.1.1

Tulis kembali menggunakan sifat komutatif dari perkalian.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 \cdot (5 x^{\frac{1}{5}} x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.2

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{1}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.3.3.1.2.1

Pindahkan $x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 \cdot (5 (x^{\frac{1}{5}} x^{\frac{1}{5}})) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.2.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 \cdot (5 x^{\frac{1}{5} + \frac{1}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.2.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 \cdot (5 x^{\frac{1 + 1}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.2.4

Tambahkan $1$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (5 \cdot (5 x^{\frac{2}{5}}) + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.3

Kalikan $5$ dengan $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}} + 5 x^{\frac{1}{5}} \cdot - 1 - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.4

Kalikan $- 1$ dengan $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}} - 5 x^{\frac{1}{5}} - 1 (5 x^{\frac{1}{5}}) - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.5

Kalikan $5$ dengan $- 1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}} - 5 x^{\frac{1}{5}} - 5 x^{\frac{1}{5}} - 1 \cdot - 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.1.6

Kalikan $- 1$ dengan $- 1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}} - 5 x^{\frac{1}{5}} - 5 x^{\frac{1}{5}} + 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.3.2

Kurangi $5 x^{\frac{1}{5}}$ dengan $- 5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} + 1) + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.4

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 2 (25 x^{\frac{2}{5}}) + 2 (- 10 x^{\frac{1}{5}}) + 2 \cdot 1 + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.5

Sederhanakan.

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Langkah 5.2.3.3.5.1

Kalikan $25$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} + 2 (- 10 x^{\frac{1}{5}}) + 2 \cdot 1 + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.5.2

Kalikan $- 10$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 \cdot 1 + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.5.3

Kalikan $2$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 {(5 x^{\frac{1}{5}} - 1)}^{3} + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.6

Gunakan Teorema Binomial.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 ({(5 x^{\frac{1}{5}})}^{3} + 3 {(5 x^{\frac{1}{5}})}^{2} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7

Sederhanakan setiap suku.

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Langkah 5.2.3.3.7.1

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (5^{3} {(x^{\frac{1}{5}})}^{3} + 3 {(5 x^{\frac{1}{5}})}^{2} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.2

Naikkan $5$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 {(x^{\frac{1}{5}})}^{3} + 3 {(5 x^{\frac{1}{5}})}^{2} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.3

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{3}$ .

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Langkah 5.2.3.3.7.3.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{1}{5} \cdot 3} + 3 {(5 x^{\frac{1}{5}})}^{2} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.3.2

Gabungkan $\frac{1}{5}$ dan $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 3 {(5 x^{\frac{1}{5}})}^{2} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.4

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 3 (5^{2} {(x^{\frac{1}{5}})}^{2}) \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.5

Naikkan $5$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 3 (25 {(x^{\frac{1}{5}})}^{2}) \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.6

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{2}$ .

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Langkah 5.2.3.3.7.6.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 3 (25 x^{\frac{1}{5} \cdot 2}) \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.6.2

Gabungkan $\frac{1}{5}$ dan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 3 (25 x^{\frac{2}{5}}) \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.7

Kalikan $25$ dengan $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} + 75 x^{\frac{2}{5}} \cdot - 1 + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.8

Kalikan $- 1$ dengan $75$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} - 75 x^{\frac{2}{5}} + 3 (5 x^{\frac{1}{5}}) {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.9

Kalikan $5$ dengan $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} - 75 x^{\frac{2}{5}} + 15 x^{\frac{1}{5}} {(- 1)}^{2} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.10

Naikkan $- 1$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} - 75 x^{\frac{2}{5}} + 15 x^{\frac{1}{5}} \cdot 1 + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.11

Kalikan $15$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} - 75 x^{\frac{2}{5}} + 15 x^{\frac{1}{5}} + {(- 1)}^{3}) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.7.12

Naikkan $- 1$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}} - 75 x^{\frac{2}{5}} + 15 x^{\frac{1}{5}} - 1) + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.8

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 2 (125 x^{\frac{3}{5}}) + 2 (- 75 x^{\frac{2}{5}}) + 2 (15 x^{\frac{1}{5}}) + 2 \cdot - 1 + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.9

Sederhanakan.

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Langkah 5.2.3.3.9.1

Kalikan $125$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} + 2 (- 75 x^{\frac{2}{5}}) + 2 (15 x^{\frac{1}{5}}) + 2 \cdot - 1 + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.9.2

Kalikan $- 75$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 2 (15 x^{\frac{1}{5}}) + 2 \cdot - 1 + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.9.3

Kalikan $15$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} + 2 \cdot - 1 + {(5 x^{\frac{1}{5}} - 1)}^{4}}{625}$

Langkah 5.2.3.3.9.4

Kalikan $2$ dengan $- 1$ .

Langkah 5.2.3.3.10

Gunakan Teorema Binomial.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + {(5 x^{\frac{1}{5}})}^{4} + 4 {(5 x^{\frac{1}{5}})}^{3} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11

Sederhanakan setiap suku.

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Langkah 5.2.3.3.11.1

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 5^{4} {(x^{\frac{1}{5}})}^{4} + 4 {(5 x^{\frac{1}{5}})}^{3} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.2

Naikkan $5$ menjadi pangkat $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 {(x^{\frac{1}{5}})}^{4} + 4 {(5 x^{\frac{1}{5}})}^{3} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.3

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{4}$ .

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Langkah 5.2.3.3.11.3.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{1}{5} \cdot 4} + 4 {(5 x^{\frac{1}{5}})}^{3} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.3.2

Gabungkan $\frac{1}{5}$ dan $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 4 {(5 x^{\frac{1}{5}})}^{3} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.4

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 4 (5^{3} {(x^{\frac{1}{5}})}^{3}) \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.5

Naikkan $5$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 4 (125 {(x^{\frac{1}{5}})}^{3}) \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.6

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{3}$ .

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Langkah 5.2.3.3.11.6.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 4 (125 x^{\frac{1}{5} \cdot 3}) \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.6.2

Gabungkan $\frac{1}{5}$ dan $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 4 (125 x^{\frac{3}{5}}) \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.7

Kalikan $125$ dengan $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} + 500 x^{\frac{3}{5}} \cdot - 1 + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.8

Kalikan $- 1$ dengan $500$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 6 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.9

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 6 (5^{2} {(x^{\frac{1}{5}})}^{2}) {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.10

Naikkan $5$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 6 (25 {(x^{\frac{1}{5}})}^{2}) {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.11

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{2}$ .

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Langkah 5.2.3.3.11.11.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 6 (25 x^{\frac{1}{5} \cdot 2}) {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.11.2

Gabungkan $\frac{1}{5}$ dan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 6 (25 x^{\frac{2}{5}}) {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.12

Kalikan $25$ dengan $6$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} {(- 1)}^{2} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.13

Naikkan $- 1$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} \cdot 1 + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.14

Kalikan $150$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} + 4 (5 x^{\frac{1}{5}}) {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.15

Kalikan $5$ dengan $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} + 20 x^{\frac{1}{5}} {(- 1)}^{3} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.16

Naikkan $- 1$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} + 20 x^{\frac{1}{5}} \cdot - 1 + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.17

Kalikan $- 1$ dengan $20$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + {(- 1)}^{4}}{625}$

Langkah 5.2.3.3.11.18

Naikkan $- 1$ menjadi pangkat $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 1}{625}$

Langkah 5.2.3.4

Sederhanakan dengan menambahkan suku-suku.

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Langkah 5.2.3.4.1

Gabungkan suku balikan dalam $5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 2 + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} - 2 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 1$ .

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Langkah 5.2.3.4.1.1

Kurangi $2$ dengan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} + 0 + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 1}{625}$

Langkah 5.2.3.4.1.2

Tambahkan $5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} - 150 x^{\frac{2}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 150 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 1}{625}$

Langkah 5.2.3.4.1.3

Tambahkan $- 150 x^{\frac{2}{5}}$ dan $150 x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} + 0 - 20 x^{\frac{1}{5}} + 1}{625}$

Langkah 5.2.3.4.1.4

Tambahkan $5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} - 1 + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}} + 1}{625}$

Langkah 5.2.3.4.1.5

Tambahkan $- 1$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}} + 0}{625}$

Langkah 5.2.3.4.1.6

Tambahkan $5 x^{\frac{1}{5}} + 50 x^{\frac{2}{5}} - 20 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}}$ dan $0$ .

Langkah 5.2.3.4.2

Kurangi $20 x^{\frac{1}{5}}$ dengan $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{50 x^{\frac{2}{5}} - 15 x^{\frac{1}{5}} + 250 x^{\frac{3}{5}} + 30 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}}}{625}$

Langkah 5.2.3.4.3

Tambahkan $- 15 x^{\frac{1}{5}}$ dan $30 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{50 x^{\frac{2}{5}} + 250 x^{\frac{3}{5}} + 15 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 500 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}}}{625}$

Langkah 5.2.3.4.4

Kurangi $500 x^{\frac{3}{5}}$ dengan $250 x^{\frac{3}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{50 x^{\frac{2}{5}} + 15 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 250 x^{\frac{3}{5}} - 20 x^{\frac{1}{5}}}{625}$

Langkah 5.2.3.4.5

Kurangi $20 x^{\frac{1}{5}}$ dengan $15 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{50 x^{\frac{2}{5}} - 5 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 250 x^{\frac{3}{5}}}{625}$

Langkah 5.2.3.5

Sederhanakan setiap suku.

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Langkah 5.2.3.5.1

Sederhanakan pembilangnya.

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Langkah 5.2.3.5.1.1

Faktorkan $5 x^{\frac{1}{5}}$ dari $50 x^{\frac{2}{5}} - 5 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 250 x^{\frac{3}{5}}$ .

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Langkah 5.2.3.5.1.1.1

Faktorkan $5 x^{\frac{1}{5}}$ dari $50 x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}}) - 5 x^{\frac{1}{5}} + 625 x^{\frac{4}{5}} - 250 x^{\frac{3}{5}}}{625}$

Langkah 5.2.3.5.1.1.2

Faktorkan $5 x^{\frac{1}{5}}$ dari $- 5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} (- 1) + 625 x^{\frac{4}{5}} - 250 x^{\frac{3}{5}}}{625}$

Langkah 5.2.3.5.1.1.3

Faktorkan $5 x^{\frac{1}{5}}$ dari $625 x^{\frac{4}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} (- 1) + 5 x^{\frac{1}{5}} (125 x^{\frac{3}{5}}) - 250 x^{\frac{3}{5}}}{625}$

Langkah 5.2.3.5.1.1.4

Faktorkan $5 x^{\frac{1}{5}}$ dari $- 250 x^{\frac{3}{5}}$ .

Langkah 5.2.3.5.1.1.5

Faktorkan $5 x^{\frac{1}{5}}$ dari $5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}}) + 5 x^{\frac{1}{5}} (- 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}} - 1) + 5 x^{\frac{1}{5}} (125 x^{\frac{3}{5}}) + 5 x^{\frac{1}{5}} (- 50 x^{\frac{2}{5}})}{625}$

Langkah 5.2.3.5.1.1.6

Faktorkan $5 x^{\frac{1}{5}}$ dari $5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}} - 1) + 5 x^{\frac{1}{5}} (125 x^{\frac{3}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}} - 1 + 125 x^{\frac{3}{5}}) + 5 x^{\frac{1}{5}} (- 50 x^{\frac{2}{5}})}{625}$

Langkah 5.2.3.5.1.1.7

Faktorkan $5 x^{\frac{1}{5}}$ dari $5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}} - 1 + 125 x^{\frac{3}{5}}) + 5 x^{\frac{1}{5}} (- 50 x^{\frac{2}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (10 x^{\frac{1}{5}} - 1 + 125 x^{\frac{3}{5}} - 50 x^{\frac{2}{5}})}{625}$

Langkah 5.2.3.5.1.2

Susun kembali suku-suku.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{625}$

Langkah 5.2.3.5.2

Faktorkan $5$ dari $5 x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 (x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1))}{625}$

Langkah 5.2.3.5.3

Batalkan faktor persekutuan.

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Langkah 5.2.3.5.3.1

Faktorkan $5$ dari $625$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{5 (x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1))}{5 \cdot 125}$

Langkah 5.2.3.5.3.2

Batalkan faktor persekutuan.

Langkah 5.2.3.5.3.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1}{3125} + \frac{{(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.6

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + {(5 x^{\frac{1}{5}} - 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7

Sederhanakan setiap suku.

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Langkah 5.2.3.7.1

Sederhanakan pembilangnya.

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Langkah 5.2.3.7.1.1

Gunakan Teorema Binomial.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + {(5 x^{\frac{1}{5}})}^{5} + 5 {(5 x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2

Sederhanakan setiap suku.

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Langkah 5.2.3.7.1.2.1

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 5^{5} {(x^{\frac{1}{5}})}^{5} + 5 {(5 x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.2

Naikkan $5$ menjadi pangkat $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 {(x^{\frac{1}{5}})}^{5} + 5 {(5 x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.3

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{5}$ .

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Langkah 5.2.3.7.1.2.3.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x^{\frac{1}{5} \cdot 5} + 5 {(5 x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.3.2

Batalkan faktor persekutuan dari $5$ .

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Langkah 5.2.3.7.1.2.3.2.1

Batalkan faktor persekutuan.

Langkah 5.2.3.7.1.2.3.2.2

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 5 {(5 x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.4

Sederhanakan.

Langkah 5.2.3.7.1.2.5

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 5 (5^{4} {(x^{\frac{1}{5}})}^{4}) \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.6

Kalikan $5$ dengan $5^{4}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.7.1.2.6.1

Pindahkan $5^{4}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 5^{4} \cdot (5 {(x^{\frac{1}{5}})}^{4}) \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.6.2

Kalikan $5^{4}$ dengan $5$ .

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Langkah 5.2.3.7.1.2.6.2.1

Naikkan $5$ menjadi pangkat $1$ .

Langkah 5.2.3.7.1.2.6.2.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 5^{4 + 1} {(x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.6.3

Tambahkan $4$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 5^{5} {(x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.7

Naikkan $5$ menjadi pangkat $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 3125 {(x^{\frac{1}{5}})}^{4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.8

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{4}$ .

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Langkah 5.2.3.7.1.2.8.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 3125 x^{\frac{1}{5} \cdot 4} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.8.2

Gabungkan $\frac{1}{5}$ dan $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x + 3125 x^{\frac{4}{5}} \cdot - 1 + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.9

Kalikan $- 1$ dengan $3125$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 10 {(5 x^{\frac{1}{5}})}^{3} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.10

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 10 (5^{3} {(x^{\frac{1}{5}})}^{3}) {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.11

Naikkan $5$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 10 (125 {(x^{\frac{1}{5}})}^{3}) {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.12

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{3}$ .

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Langkah 5.2.3.7.1.2.12.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 10 (125 x^{\frac{1}{5} \cdot 3}) {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.12.2

Gabungkan $\frac{1}{5}$ dan $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 10 (125 x^{\frac{3}{5}}) {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.13

Kalikan $125$ dengan $10$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} {(- 1)}^{2} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.14

Naikkan $- 1$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} \cdot 1 + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.15

Kalikan $1250$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 10 {(5 x^{\frac{1}{5}})}^{2} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.16

Terapkan kaidah hasil kali ke $5 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 10 (5^{2} {(x^{\frac{1}{5}})}^{2}) {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.17

Naikkan $5$ menjadi pangkat $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 10 (25 {(x^{\frac{1}{5}})}^{2}) {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.18

Kalikan eksponen dalam ${(x^{\frac{1}{5}})}^{2}$ .

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Langkah 5.2.3.7.1.2.18.1

Terapkan kaidah pangkat dan perkalian eksponen, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 10 (25 x^{\frac{1}{5} \cdot 2}) {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.18.2

Gabungkan $\frac{1}{5}$ dan $2$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 10 (25 x^{\frac{2}{5}}) {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.19

Kalikan $25$ dengan $10$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 250 x^{\frac{2}{5}} {(- 1)}^{3} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.20

Naikkan $- 1$ menjadi pangkat $3$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} + 250 x^{\frac{2}{5}} \cdot - 1 + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.21

Kalikan $- 1$ dengan $250$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 5 (5 x^{\frac{1}{5}}) {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.22

Kalikan $5$ dengan $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}} {(- 1)}^{4} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.23

Naikkan $- 1$ menjadi pangkat $4$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}} \cdot 1 + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.24

Kalikan $25$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}} + {(- 1)}^{5}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.2.25

Naikkan $- 1$ menjadi pangkat $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{1 + 3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}} - 1}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.3

Kurangi $1$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}} + 0}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.4

Tambahkan $3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{3125 x - 3125 x^{\frac{4}{5}} + 1250 x^{\frac{3}{5}} - 250 x^{\frac{2}{5}} + 25 x^{\frac{1}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.5

Susun kembali suku-suku.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{3125 x + 1250 x^{\frac{3}{5}} + 25 x^{\frac{1}{5}} - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6

Tulis kembali $3125 x + 1250 x^{\frac{3}{5}} + 25 x^{\frac{1}{5}} - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}$ dalam bentuk faktor.

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Langkah 5.2.3.7.1.6.1

Faktorkan $25 x^{\frac{1}{5}}$ dari $3125 x + 1250 x^{\frac{3}{5}} + 25 x^{\frac{1}{5}} - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}$ .

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Langkah 5.2.3.7.1.6.1.1

Faktorkan $25 x^{\frac{1}{5}}$ dari $3125 x$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 1250 x^{\frac{3}{5}} + 25 x^{\frac{1}{5}} - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.2

Faktorkan $25 x^{\frac{1}{5}}$ dari $1250 x^{\frac{3}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 25 x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.3

Faktorkan $25 x^{\frac{1}{5}}$ dari $25 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 25 x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} (1) - 250 x^{\frac{2}{5}} - 3125 x^{\frac{4}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.4

Faktorkan $25 x^{\frac{1}{5}}$ dari $- 250 x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 25 x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} (1) + 25 x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) - 3125 x^{\frac{4}{5}}}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.5

Faktorkan $25 x^{\frac{1}{5}}$ dari $- 3125 x^{\frac{4}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 25 x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} (1) + 25 x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + 25 x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.6

Faktorkan $25 x^{\frac{1}{5}}$ dari $25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 25 x^{\frac{1}{5}} (50 x^{\frac{2}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} (1) + 25 x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + 25 x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.7

Faktorkan $25 x^{\frac{1}{5}}$ dari $25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}}) + 25 x^{\frac{1}{5}} (1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} + 1) + 25 x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + 25 x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.8

Faktorkan $25 x^{\frac{1}{5}}$ dari $25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} + 1) + 25 x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} + 1 - 10 x^{\frac{1}{5}}) + 25 x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.1.9

Faktorkan $25 x^{\frac{1}{5}}$ dari $25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} + 1 - 10 x^{\frac{1}{5}}) + 25 x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} + 1 - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}})}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.1.6.2

Susun kembali suku-suku.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1)}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.2

Faktorkan $25$ dari $25 x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 (x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1))}{3125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.3

Batalkan faktor persekutuan.

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Langkah 5.2.3.7.3.1

Faktorkan $25$ dari $3125$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{25 (x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1))}{25 \cdot 125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.7.3.2

Batalkan faktor persekutuan.

Langkah 5.2.3.7.3.3

Tulis kembali pernyataannya.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1)}{125} + \frac{x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.8

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1) + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9

Sederhanakan setiap suku.

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Langkah 5.2.3.9.1

Terapkan sifat distributif.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.2

Sederhanakan.

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Langkah 5.2.3.9.2.1

Tulis kembali menggunakan sifat komutatif dari perkalian.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{1}{5}} x^{\frac{4}{5}} + x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.2.2

Tulis kembali menggunakan sifat komutatif dari perkalian.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{1}{5}} x^{\frac{4}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.2.3

Tulis kembali menggunakan sifat komutatif dari perkalian.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{1}{5}} x^{\frac{4}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.2.4

Tulis kembali menggunakan sifat komutatif dari perkalian.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{1}{5}} x^{\frac{4}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.2.5

Kalikan $x^{\frac{1}{5}}$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{1}{5}} x^{\frac{4}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3

Sederhanakan setiap suku.

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Langkah 5.2.3.9.3.1

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{4}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.9.3.1.1

Pindahkan $x^{\frac{4}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 (x^{\frac{4}{5}} x^{\frac{1}{5}}) + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.1.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{4}{5} + \frac{1}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.1.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{4 + 1}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.1.4

Tambahkan $4$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x^{\frac{5}{5}} + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.1.5

Bagilah $5$ dengan $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{1}{5}} x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.2

Sederhanakan $125 x^{1}$ .

Langkah 5.2.3.9.3.3

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{2}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.9.3.3.1

Pindahkan $x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 (x^{\frac{2}{5}} x^{\frac{1}{5}}) - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.3.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{2}{5} + \frac{1}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.3.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{2 + 1}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.3.4

Tambahkan $2$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{1}{5}} x^{\frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.4

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{1}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.9.3.4.1

Pindahkan $x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 (x^{\frac{1}{5}} x^{\frac{1}{5}}) - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.4.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{1}{5} + \frac{1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.4.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{1 + 1}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.4.4

Tambahkan $1$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{1}{5}} x^{\frac{3}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.5

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{3}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.3.9.3.5.1

Pindahkan $x^{\frac{3}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 (x^{\frac{3}{5}} x^{\frac{1}{5}}) + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.5.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{3}{5} + \frac{1}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.5.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{3 + 1}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.3.9.3.5.4

Tambahkan $3$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{4}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4

Sederhanakan pembilangnya.

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Langkah 5.2.4.1

Faktorkan $x^{\frac{1}{5}}$ dari $125 x + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{4}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)$ .

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Langkah 5.2.4.1.1

Faktorkan $x^{\frac{1}{5}}$ dari $125 x$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + 50 x^{\frac{3}{5}} - 10 x^{\frac{2}{5}} - 125 x^{\frac{4}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.2

Faktorkan $x^{\frac{1}{5}}$ dari $50 x^{\frac{3}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) - 10 x^{\frac{2}{5}} - 125 x^{\frac{4}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.3

Faktorkan $x^{\frac{1}{5}}$ dari $- 10 x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) - 125 x^{\frac{4}{5}} + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.4

Faktorkan $x^{\frac{1}{5}}$ dari $- 125 x^{\frac{4}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + x^{\frac{1}{5}} (50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.5

Kalikan dengan $1$ .

Langkah 5.2.4.1.6

Faktorkan $x^{\frac{1}{5}}$ dari $x^{\frac{1}{5}} (125 x^{\frac{4}{5}}) + x^{\frac{1}{5}} (50 x^{\frac{2}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.7

Faktorkan $x^{\frac{1}{5}}$ dari $x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}}) + x^{\frac{1}{5}} (- 10 x^{\frac{1}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.8

Faktorkan $x^{\frac{1}{5}}$ dari $x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}}) + x^{\frac{1}{5}} (- 125 x^{\frac{3}{5}})$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1 + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.1.9

Faktorkan $x^{\frac{1}{5}}$ dari $x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}}) + x^{\frac{1}{5}} \cdot 1$ .

Langkah 5.2.4.1.10

Faktorkan $x^{\frac{1}{5}}$ dari $x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1) + x^{\frac{1}{5}} (125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 50 x^{\frac{2}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1 + 125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 50 x^{\frac{2}{5}} - 1)}{125}$

Langkah 5.2.4.2

Kurangi $50 x^{\frac{2}{5}}$ dengan $50 x^{\frac{2}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1 + 125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} + 0 - 1)}{125}$

Langkah 5.2.4.3

Tambahkan $125 x^{\frac{4}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} - 10 x^{\frac{1}{5}} - 125 x^{\frac{3}{5}} + 1 + 125 x^{\frac{3}{5}} + 10 x^{\frac{1}{5}} - 1)}{125}$

Langkah 5.2.4.4

Tambahkan $- 10 x^{\frac{1}{5}}$ dan $10 x^{\frac{1}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} - 125 x^{\frac{3}{5}} + 1 + 125 x^{\frac{3}{5}} + 0 - 1)}{125}$

Langkah 5.2.4.5

Tambahkan $125 x^{\frac{4}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} - 125 x^{\frac{3}{5}} + 1 + 125 x^{\frac{3}{5}} - 1)}{125}$

Langkah 5.2.4.6

Tambahkan $- 125 x^{\frac{3}{5}}$ dan $125 x^{\frac{3}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 0 + 1 - 1)}{125}$

Langkah 5.2.4.7

Tambahkan $125 x^{\frac{4}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 1 - 1)}{125}$

Langkah 5.2.4.8

Kurangi $1$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} (125 x^{\frac{4}{5}} + 0)}{125}$

Langkah 5.2.4.9

Tambahkan $125 x^{\frac{4}{5}}$ dan $0$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{1}{5}} \cdot (125 x^{\frac{4}{5}})}{125}$

Langkah 5.2.4.10

Gabungkan eksponen.

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Langkah 5.2.4.10.1

Kalikan $x^{\frac{1}{5}}$ dengan $x^{\frac{4}{5}}$ dengan menambahkan eksponennya.

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Langkah 5.2.4.10.1.1

Pindahkan $x^{\frac{4}{5}}$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{4}{5}} x^{\frac{1}{5}} \cdot 125}{125}$

Langkah 5.2.4.10.1.2

Gunakan kaidah pangkat $a^{m} a^{n} = a^{m + n}$ untuk menggabungkan pangkat.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{4}{5} + \frac{1}{5}} \cdot 125}{125}$

Langkah 5.2.4.10.1.3

Gabungkan pembilang dari penyebut persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{4 + 1}{5}} \cdot 125}{125}$

Langkah 5.2.4.10.1.4

Tambahkan $4$ dan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x^{\frac{5}{5}} \cdot 125}{125}$

Langkah 5.2.4.10.1.5

Bagilah $5$ dengan $5$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x \cdot 125}{125}$

Langkah 5.2.4.10.2

Sederhanakan $x^{1} \cdot 125$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x \cdot 125}{125}$

Langkah 5.2.5

Batalkan faktor persekutuan dari $125$ .

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Langkah 5.2.5.1

Batalkan faktor persekutuan.

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = \frac{x \cdot 125}{125}$

Langkah 5.2.5.2

Bagilah $x$ dengan $1$ .

$f^{- 1} (10 x^{\frac{1}{5}} - 2) = x$

Langkah 5.3

Evaluasi $f (f^{- 1} (x))$ .

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Langkah 5.3.1

Tulis fungsi hasil komposit.

$f (f^{- 1} (x))$

Langkah 5.3.2

Evaluasi $f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000})$ dengan mensubstitusikan nilai (Variabel1) ke dalam (Variabel2).

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000})}^{\frac{1}{5}} - 2$

Langkah 5.3.3

Pindahkan $\frac{1}{3125}$ .

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000} + \frac{1}{3125})}^{\frac{1}{5}} - 2$

Langkah 5.3.4

Pindahkan $\frac{x}{1250}$ .

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000} + \frac{x}{1250} + \frac{1}{3125})}^{\frac{1}{5}} - 2$

Langkah 5.3.5

Pindahkan $\frac{x^{2}}{1250}$ .

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000} + \frac{x^{2}}{1250} + \frac{x}{1250} + \frac{1}{3125})}^{\frac{1}{5}} - 2$

Langkah 5.3.6

Pindahkan $\frac{x^{3}}{2500}$ .

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{x^{4}}{10000} + \frac{x^{5}}{100000} + \frac{x^{3}}{2500} + \frac{x^{2}}{1250} + \frac{x}{1250} + \frac{1}{3125})}^{\frac{1}{5}} - 2$

Langkah 5.3.7

Susun kembali $\frac{x^{4}}{10000}$ dan $\frac{x^{5}}{100000}$ .

$f (\frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}) = 10 {(\frac{x^{5}}{100000} + \frac{x^{4}}{10000} + \frac{x^{3}}{2500} + \frac{x^{2}}{1250} + \frac{x}{1250} + \frac{1}{3125})}^{\frac{1}{5}} - 2$

Langkah 5.4

Karena $f^{- 1} (f (x)) = x$ dan $f (f^{- 1} (x)) = x$ , maka $f^{- 1} (x) = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}$ merupakan balikan dari $f (x) = 10 x^{\frac{1}{5}} - 2$ .

$f^{- 1} (x) = \frac{1}{3125} + \frac{x}{1250} + \frac{x^{2}}{1250} + \frac{x^{3}}{2500} + \frac{x^{4}}{10000} + \frac{x^{5}}{100000}$