Aljabar Contoh

${(\frac{\sqrt{x}}{y^{2}} - \frac{y}{x})}^{18}$

Langkah 1

Gunakan teorema pengembangan binomial untuk menentukan setiap suku. Teorema binomial menyatakan ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{18} \frac{18!}{(18 - k)! k!} \cdot {(\frac{\sqrt{x}}{y^{2}})}^{18 - k} \cdot {(- \frac{y}{x})}^{k}$

Langkah 2

Perluas penjumlahannya.

$\frac{18!}{(18 - 0)! 0!} \cdot {(\frac{\sqrt{x}}{y^{2}})}^{18 - 0} \cdot {(- \frac{y}{x})}^{0} + \frac{18!}{(18 - 1)! 1!} \cdot {(\frac{\sqrt{x}}{y^{2}})}^{18 - 1} \cdot {(- \frac{y}{x})}^{1} + \dots + \frac{18!}{(18 - 18)! 18!} \cdot {(\frac{\sqrt{x}}{y^{2}})}^{18 - 18} \cdot {(- \frac{y}{x})}^{18}$

Langkah 3

Sederhanakan eksponen untuk setiap suku dari pengembangan.

$1 \cdot {(\frac{\sqrt{x}}{y^{2}})}^{18} \cdot {(- \frac{y}{x})}^{0} + 18 \cdot {(\frac{\sqrt{x}}{y^{2}})}^{17} \cdot {(- \frac{y}{x})}^{1} + \dots + 1 \cdot {(\frac{\sqrt{x}}{y^{2}})}^{0} \cdot {(- \frac{y}{x})}^{18}$

Langkah 4

Sederhanakan hasil polinomialnya.

$\frac{x^{9}}{y^{36}} - \frac{18 x^{7} \sqrt{x}}{y^{33}} + \frac{153 x^{6}}{y^{30}} - \frac{816 x^{4} \sqrt{x}}{y^{27}} + \frac{3060 x^{3}}{y^{24}} - \frac{8568 x \sqrt{x}}{y^{21}} + \frac{18564}{y^{18}} - \frac{31824 \sqrt{x}}{y^{15} x^{2}} + \frac{43758}{y^{12} x^{3}} - \frac{48620 \sqrt{x}}{y^{9} x^{5}} + \frac{43758}{y^{6} x^{6}} - \frac{31824 \sqrt{x}}{y^{3} x^{8}} + \frac{18564}{x^{9}} - \frac{8568 y^{3} \sqrt{x}}{x^{11}} + \frac{3060 y^{6}}{x^{12}} - \frac{816 y^{9} \sqrt{x}}{x^{14}} + \frac{153 y^{12}}{x^{15}} - \frac{18 y^{15} \sqrt{x}}{x^{17}} + \frac{y^{18}}{x^{18}}$