第127页

信息发布者:
$5$
$\frac{1}{3}$
$\frac{3\sqrt{3}}{2}$
$2-\sqrt{3}$或$-2+\sqrt{3}$
解:原式​$=4\sqrt {3}×(12\sqrt {3}-\frac {\sqrt {2}}{3}-9\sqrt {3})$​
​                $=4\sqrt {3}×(3\sqrt {3}-\frac {\sqrt {2}}{3})$​
​                $=4\sqrt {3}×3\sqrt {3}-4\sqrt {3}×\frac {\sqrt {2}}{3}$​
​                $=36-\frac {4\sqrt {6}}{3}$​
解:原式​$=5\sqrt {48}÷\sqrt {3}+\sqrt {12}÷\sqrt {3}-7\sqrt {7}÷\sqrt {3}$​
​                $=5\sqrt {16}+\sqrt {4}-\frac {7\sqrt {21}}{3}$​
​                $=20+2-\frac {7\sqrt {21}}{3}$​
​                $=22-\frac {7\sqrt {21}}{3}$​
解:原式​$=\frac {(\sqrt {5})^2-2×\sqrt {5}×2+2^2}{9}$​
​                $=\frac {5-4\sqrt {5}+4}{9}$​
​                $=\frac {9-4\sqrt {5}}{9}$​
解:原式​$=(2\sqrt {5})^2-(5\sqrt {2})^2-(5-2\sqrt {10}+2)$​
​                $=20-50-7+2\sqrt {10}$​
​                $=2\sqrt {10}-37$​
解:
$\begin{aligned}原式&=\frac{2m(m+2)}{m-2}·\frac{(m-2)^2}{m}\\&=2(m+2)(m-2)\\&=2(m^2-4)\\&=2m^2-8\end{aligned}$
当$m=\sqrt{3}-1$时,
$\begin{aligned}原式&=2(\sqrt{3}-1)^2-8\\&=2(3-2\sqrt{3}+1)-8\\&=8-4\sqrt{3}-8\\&=-4\sqrt{3}\end{aligned}$
情况1:选择$-\sqrt{2}$和$\sqrt{3}$
解:
$\begin{aligned}(-\sqrt{2}+\sqrt{3})^2÷\sqrt{2}&=(5-2\sqrt{6})÷\sqrt{2}\\&=\frac{5\sqrt{2}}{2}-2\sqrt{3}\end{aligned}$
情况2:选择$-\sqrt{2}$和$\sqrt{6}$
解:
$\begin{aligned}(-\sqrt{2}+\sqrt{6})^2÷\sqrt{2}&=(8-2\sqrt{12})÷\sqrt{2}\\&=4\sqrt{2}-2\sqrt{6}\end{aligned}$
情况3:选择$\sqrt{3}$和$\sqrt{6}$
解:
$\begin{aligned}(\sqrt{3}+\sqrt{6})^2÷\sqrt{2}&=(9+2\sqrt{18})÷\sqrt{2}\\&=\frac{9\sqrt{2}}{2}+6\end{aligned}$