第122页

信息发布者：
C
8
$-\frac{2\sqrt{5}}{5}$
-2
解：原式​$=\sqrt {\frac {49}{9}a²×a}$​
​$=\frac {7a}{3}\sqrt {a}$​
$ \begin{aligned}解:原式&=\sqrt {\frac {9}{36}-\frac {4}{36}} \\ &=\sqrt {\frac {5}{36}} \\ &=\frac{\sqrt{5}}{6} \\ \end{aligned}$
$ \begin{aligned}解:原式&=\frac {(m-n)×\sqrt {m-n}}{\sqrt {m-n}×\sqrt {m-n}} \\ &=\sqrt {m-n} \\ \end{aligned}$
解：原式​$=\sqrt {x^4×\frac {2}{x}}$​
​$=\sqrt {x²×2x}$​
​$=x\sqrt {2x}$​
解：原式​$=-12\sqrt {48}÷12\sqrt {3}$​
​$=-\sqrt {16}$​
​$=-4$​
$ \begin{aligned} 解:原式&=-\sqrt {10}÷(-2\sqrt 5)×(-\frac {\sqrt 2}{2}) \\ &=\frac {\sqrt 2}{2}×(-\frac {\sqrt 2}{2}) \\ &=-\frac{1}{2} \\ \end{aligned}$
解：原式$=(\frac{x + 3}{x^{2}-1}-\frac{2x + 2}{x^{2}-1})\cdot\frac{x(x + 1)}{x + 2}$
$=\frac{-(x - 1)}{(x + 1)(x - 1)}\cdot\frac{x(x + 1)}{x + 2}=-\frac{x}{x + 2}。$
当$x = \sqrt{2}-2$时，
原式$=-\frac{\sqrt{2}-2}{\sqrt{2}-2 + 2}=-\frac{\sqrt{2}-2}{\sqrt{2}}=\sqrt{2}-1。$
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