代数示例

$\ln (\sqrt{\frac{c^{2} - z^{2}}{c^{2} + z^{2}}})$

解题步骤 1

使用 $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ ，将 $\sqrt{\frac{c^{2} - z^{2}}{c^{2} + z^{2}}}$ 重写成 ${(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}$ 。

$\frac{d}{d c} [\ln ({(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}})]$

解题步骤 2

使用链式法则求微分，根据该法则， $\frac{d}{d c} [f (g (c))]$ 等于 $f' (g (c)) g' (c)$ ，其中 $f (c) = \ln (c)$ 且 $g (c) = {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}$ 。

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解题步骤 2.1

要使用链式法则，请将 $u_{1}$ 设为 ${(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}$ 。

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d c} [{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}]$

解题步骤 2.2

$\ln (u_{1})$ 对 $u_{1}$ 的导数为 $\frac{1}{u_{1}}$ 。

$\frac{1}{u_{1}} \frac{d}{d c} [{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}]$

解题步骤 2.3

使用 ${(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}$ 替换所有出现的 $u_{1}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} \frac{d}{d c} [{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}]$

解题步骤 3

使用链式法则求微分，根据该法则， $\frac{d}{d c} [f (g (c))]$ 等于 $f' (g (c)) g' (c)$ ，其中 $f (c) = c^{\frac{1}{2}}$ 且 $g (c) = \frac{c^{2} - z^{2}}{c^{2} + z^{2}}$ 。

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解题步骤 3.1

要使用链式法则，请将 $u_{2}$ 设为 $\frac{c^{2} - z^{2}}{c^{2} + z^{2}}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 3.2

使用幂法则求微分，根据该法则， $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ 等于 $n {u_{2}}^{n - 1}$ ，其中 $n = \frac{1}{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 3.3

使用 $\frac{c^{2} - z^{2}}{c^{2} + z^{2}}$ 替换所有出现的 $u_{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2} - 1} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 4

要将 $- 1$ 写成带有公分母的分数，请乘以 $\frac{2}{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 5

组合 $- 1$ 和 $\frac{2}{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 6

在公分母上合并分子。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 7

化简分子。

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解题步骤 7.1

将 $- 1$ 乘以 $2$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1 - 2}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 7.2

从 $1$ 中减去 $2$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{- 1}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 8

将负号移到分数的前面。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{d}{d c} [\frac{c^{2} - z^{2}}{c^{2} + z^{2}}])$

解题步骤 9

使用除法定则求微分，根据该法则， $\frac{d}{d c} [\frac{f (c)}{g (c)}]$ 等于 $\frac{g (c) \frac{d}{d c} [f (c)] - f (c) \frac{d}{d c} [g (c)]}{{g (c)}^{2}}$ ，其中 $f (c) = c^{2} - z^{2}$ 且 $g (c) = c^{2} + z^{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{(c^{2} + z^{2}) \frac{d}{d c} [c^{2} - z^{2}] - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10

求微分。

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解题步骤 10.1

根据加法法则， $c^{2} - z^{2}$ 对 $c$ 的导数是 $\frac{d}{d c} [c^{2}] + \frac{d}{d c} [- z^{2}]$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{(c^{2} + z^{2}) (\frac{d}{d c} [c^{2}] + \frac{d}{d c} [- z^{2}]) - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.2

使用幂法则求微分，根据该法则， $\frac{d}{d c} [c^{n}]$ 等于 $n c^{n - 1}$ ，其中 $n = 2$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{(c^{2} + z^{2}) (2 c + \frac{d}{d c} [- z^{2}]) - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.3

因为 $- z^{2}$ 对于 $c$ 是常数，所以 $- z^{2}$ 对 $c$ 的导数为 $0$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{(c^{2} + z^{2}) (2 c + 0) - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.4

化简表达式。

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解题步骤 10.4.1

将 $2 c$ 和 $0$ 相加。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{(c^{2} + z^{2}) (2 c) - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.4.2

将 $2$ 移到 $c^{2} + z^{2}$ 的左侧。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{2 \cdot (c^{2} + z^{2}) c - (c^{2} - z^{2}) \frac{d}{d c} [c^{2} + z^{2}]}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.5

根据加法法则， $c^{2} + z^{2}$ 对 $c$ 的导数是 $\frac{d}{d c} [c^{2}] + \frac{d}{d c} [z^{2}]$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{2 (c^{2} + z^{2}) c - (c^{2} - z^{2}) (\frac{d}{d c} [c^{2}] + \frac{d}{d c} [z^{2}])}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.6

使用幂法则求微分，根据该法则， $\frac{d}{d c} [c^{n}]$ 等于 $n c^{n - 1}$ ，其中 $n = 2$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{2 (c^{2} + z^{2}) c - (c^{2} - z^{2}) (2 c + \frac{d}{d c} [z^{2}])}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.7

因为 $z^{2}$ 对于 $c$ 是常数，所以 $z^{2}$ 对 $c$ 的导数为 $0$ 。

解题步骤 10.8

合并分数。

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解题步骤 10.8.1

将 $2 c$ 和 $0$ 相加。

解题步骤 10.8.2

将 $2$ 乘以 $- 1$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{1}{2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}} \frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{{(c^{2} + z^{2})}^{2}})$

解题步骤 10.8.3

将 $\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{{(c^{2} + z^{2})}^{2}}$ 乘以 $\frac{1}{2}$ 。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{{(c^{2} + z^{2})}^{2} \cdot 2} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}})$

解题步骤 10.8.4

将 $2$ 移到 ${(c^{2} + z^{2})}^{2}$ 的左侧。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{2 \cdot {(c^{2} + z^{2})}^{2}} {(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{- \frac{1}{2}})$

解题步骤 11

化简。

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解题步骤 11.1

通过将底数重写为其倒数的方式改变指数的符号。

$\frac{1}{{(\frac{c^{2} - z^{2}}{c^{2} + z^{2}})}^{\frac{1}{2}}} (\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} {(\frac{c^{2} + z^{2}}{c^{2} - z^{2}})}^{\frac{1}{2}})$

解题步骤 11.2

对 $\frac{c^{2} - z^{2}}{c^{2} + z^{2}}$ 运用乘积法则。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} {(\frac{c^{2} + z^{2}}{c^{2} - z^{2}})}^{\frac{1}{2}})$

解题步骤 11.3

对 $\frac{c^{2} + z^{2}}{c^{2} - z^{2}}$ 运用乘积法则。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{2 (c^{2} + z^{2}) c - 2 (c^{2} - z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.4

运用分配律。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{(2 c^{2} + 2 z^{2}) c - 2 (c^{2} - z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.5

运用分配律。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{2 c^{2} c + 2 z^{2} c - 2 (c^{2} - z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.6

运用分配律。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{2 c^{2} c + 2 z^{2} c + (- 2 c^{2} - 2 (- z^{2})) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.7

运用分配律。

$\frac{1}{\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}} (\frac{2 c^{2} c + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8

合并项。

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解题步骤 11.8.1

乘以分数的倒数从而实现除以 $\frac{{(c^{2} - z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{\frac{1}{2}}}$ 。

$1 \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{2} c + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.2

将 $\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}}$ 乘以 $1$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{2} c + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.3

对 $c$ 进行 $1$ 次方运算。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 (c^{1} c^{2}) + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.4

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{1 + 2} + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.5

将 $1$ 和 $2$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{3} + 2 z^{2} c - 2 c^{2} c - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.6

对 $c$ 进行 $1$ 次方运算。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{3} + 2 z^{2} c - 2 (c^{1} c^{2}) - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.7

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{3} + 2 z^{2} c - 2 c^{1 + 2} - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.8

将 $1$ 和 $2$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{3} + 2 z^{2} c - 2 c^{3} - 2 (- z^{2}) c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.9

将 $- 1$ 乘以 $- 2$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 c^{3} + 2 z^{2} c - 2 c^{3} + 2 z^{2} c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.10

从 $2 c^{3}$ 中减去 $2 c^{3}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 z^{2} c + 0 + 2 z^{2} c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.11

将 $2 z^{2} c$ 和 $0$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 z^{2} c + 2 z^{2} c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.12

将 $2 z^{2} c$ 和 $2 z^{2} c$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{4 z^{2} c}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.13

约去 $4$ 和 $2$ 的公因数。

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解题步骤 11.8.13.1

从 $4 z^{2} c$ 中分解出因数 $2$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 (2 z^{2} c)}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.13.2

约去公因数。

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解题步骤 11.8.13.2.1

从 $2 {(c^{2} + z^{2})}^{2}$ 中分解出因数 $2$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 (2 z^{2} c)}{2 ({(c^{2} + z^{2})}^{2})} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.13.2.2

约去公因数。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 (2 z^{2} c)}{2 {(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.13.2.3

重写表达式。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} (\frac{2 z^{2} c}{{(c^{2} + z^{2})}^{2}} \cdot \frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}})$

解题步骤 11.8.14

将 $\frac{2 z^{2} c}{{(c^{2} + z^{2})}^{2}}$ 乘以 $\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c {(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} + z^{2})}^{2} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.15

使用负指数规则 $b^{n} = \frac{1}{b^{- n}}$ 将 ${(c^{2} + z^{2})}^{\frac{1}{2}}$ 移动到分母。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{2} {(c^{2} - z^{2})}^{\frac{1}{2}} {(c^{2} + z^{2})}^{- \frac{1}{2}}}$

解题步骤 11.8.16

通过指数相加将 ${(c^{2} + z^{2})}^{2}$ 乘以 ${(c^{2} + z^{2})}^{- \frac{1}{2}}$ 。

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解题步骤 11.8.16.1

移动 ${(c^{2} + z^{2})}^{- \frac{1}{2}}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{- \frac{1}{2}} {(c^{2} + z^{2})}^{2} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.2

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{- \frac{1}{2} + 2} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.3

要将 $2$ 写成带有公分母的分数，请乘以 $\frac{2}{2}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{- \frac{1}{2} + 2 \cdot \frac{2}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.4

组合 $2$ 和 $\frac{2}{2}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{- \frac{1}{2} + \frac{2 \cdot 2}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.5

在公分母上合并分子。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{\frac{- 1 + 2 \cdot 2}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.6

化简分子。

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解题步骤 11.8.16.6.1

将 $2$ 乘以 $2$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{\frac{- 1 + 4}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.16.6.2

将 $- 1$ 和 $4$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}} \cdot \frac{2 z^{2} c}{{(c^{2} + z^{2})}^{\frac{3}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$

解题步骤 11.8.17

将 $\frac{{(c^{2} + z^{2})}^{\frac{1}{2}}}{{(c^{2} - z^{2})}^{\frac{1}{2}}}$ 乘以 $\frac{2 z^{2} c}{{(c^{2} + z^{2})}^{\frac{3}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}}}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{\frac{1}{2}} ({(c^{2} + z^{2})}^{\frac{3}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}})}$

解题步骤 11.8.18

通过指数相加将 ${(c^{2} - z^{2})}^{\frac{1}{2}}$ 乘以 ${(c^{2} - z^{2})}^{\frac{1}{2}}$ 。

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解题步骤 11.8.18.1

移动 ${(c^{2} - z^{2})}^{\frac{1}{2}}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{\frac{1}{2}} {(c^{2} - z^{2})}^{\frac{1}{2}} {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.18.2

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{\frac{1}{2} + \frac{1}{2}} {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.18.3

在公分母上合并分子。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{\frac{1 + 1}{2}} {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.18.4

将 $1$ 和 $1$ 相加。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{\frac{2}{2}} {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.18.5

用 $2$ 除以 $2$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{{(c^{2} - z^{2})}^{1} {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.19

化简 ${(c^{2} - z^{2})}^{1} {(c^{2} + z^{2})}^{\frac{3}{2}}$ 。

$\frac{{(c^{2} + z^{2})}^{\frac{1}{2}} (2 z^{2} c)}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.20

将 $2$ 移到 ${(c^{2} + z^{2})}^{\frac{1}{2}}$ 的左侧。

$\frac{2 \cdot {(c^{2} + z^{2})}^{\frac{1}{2}} z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{\frac{3}{2}}}$

解题步骤 11.8.21

使用负指数规则 $b^{n} = \frac{1}{b^{- n}}$ 将 ${(c^{2} + z^{2})}^{\frac{1}{2}}$ 移动到分母。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{\frac{3}{2}} {(c^{2} + z^{2})}^{- \frac{1}{2}}}$

解题步骤 11.8.22

化简分母。

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解题步骤 11.8.22.1

通过指数相加将 ${(c^{2} + z^{2})}^{\frac{3}{2}}$ 乘以 ${(c^{2} + z^{2})}^{- \frac{1}{2}}$ 。

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解题步骤 11.8.22.1.1

移动 ${(c^{2} + z^{2})}^{- \frac{1}{2}}$ 。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) ({(c^{2} + z^{2})}^{- \frac{1}{2}} {(c^{2} + z^{2})}^{\frac{3}{2}})}$

解题步骤 11.8.22.1.2

使用幂法则 $a^{m} a^{n} = a^{m + n}$ 合并指数。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{- \frac{1}{2} + \frac{3}{2}}}$

解题步骤 11.8.22.1.3

在公分母上合并分子。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{\frac{- 1 + 3}{2}}}$

解题步骤 11.8.22.1.4

将 $- 1$ 和 $3$ 相加。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{\frac{2}{2}}}$

解题步骤 11.8.22.1.5

用 $2$ 除以 $2$ 。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) {(c^{2} + z^{2})}^{1}}$

解题步骤 11.8.22.2

化简 $(c^{2} - z^{2}) {(c^{2} + z^{2})}^{1}$ 。

$\frac{2 z^{2} c}{(c^{2} - z^{2}) (c^{2} + z^{2})}$

解题步骤 11.9

重新排序项。

$\frac{2 z^{2} c}{(c^{2} + z^{2}) (c^{2} - z^{2})}$

解题步骤 11.10

因为两项都是完全平方数，所以使用平方差公式 $a^{2} - b^{2} = (a + b) (a - b)$ 进行因式分解，其中 $a = c$ 和 $b = z$ 。

$\frac{2 z^{2} c}{(c^{2} + z^{2}) (c + z) (c - z)}$

代数 示例

代数示例