第31页

信息发布者：
解：原式​$=(6×\frac {\sqrt {x}}{2}-2x×\frac {\sqrt {x}}{x})÷(-\frac {\sqrt {x}}{3})$​
​$=(3\sqrt {x}-2\sqrt {x})×(-\frac {3}{\sqrt {x}})$​
​$=\sqrt {x}×(-\frac {3}{\sqrt {x}})$​
​$=-3$​
解：原式​$=(\frac {3-\sqrt {6}-1}{3})×\frac {\sqrt {6}+2}{3}$​
​$=\frac {2-\sqrt {6}}{3}×\frac {\sqrt {6}+2}{3}$​
​$=\frac {(2-\sqrt {6})(2+\sqrt {6})}{9}$​
​$=\frac {4-6}{9}$​
​$=-\frac {2}{9}$​
解：
$\begin{aligned}&(2x+y)^2+(x-y)(x+y)-5x(x-y)\\=&4x^2+4xy+y^2+x^2-y^2-5x^2+5xy\\=&9xy\end{aligned}$
当$x=\sqrt{6}-1，$$y=\sqrt{6}+1$时，
$\begin{aligned}\mathrm{原式}&=9×(\sqrt{6}-1)(\sqrt{6}+1)\\&=9×(6-1)\\&=9×5\\&=45\end{aligned}$
解：
$\begin{aligned}&(\frac{a}{a-1}-\frac{a}{a+1})÷\frac{a^2}{a^2-1}\\=&[\frac{a(a+1)-a(a-1)}{(a-1)(a+1)}]×\frac{(a-1)(a+1)}{a^2}\\=&\frac{a^2+a-a^2+a}{a^2}\\=&\frac{2a}{a^2}\\=&\frac{2}{a}\end{aligned}$
当$a=2\sqrt{3}$时，
$\begin{aligned}\mathrm{原式}&=\frac{2}{2\sqrt{3}}\\&=\frac{1}{\sqrt{3}}\\&=\frac{\sqrt{3}}{3}\end{aligned}$
解：
由题意得$\begin{cases}3x-1≥0\\1-3x≥0\end{cases}，$解得$x=\frac{1}{3}，$
则$y=2\sqrt{3×\frac{1}{3}-1}+3\sqrt{1-3×\frac{1}{3}}+\frac{1}{2}=\frac{1}{2}，$
$\begin{aligned}&\sqrt{xy}÷\frac{1}{2}\sqrt{\frac{1}{y}}\\=&\sqrt{xy}×2\sqrt{y}\\=&2\sqrt{xy^2}\\=&2y\sqrt{x}\end{aligned}$
将$x=\frac{1}{3}，$$y=\frac{1}{2}$代入得：
$\begin{aligned}\mathrm{原式}&=2×\frac{1}{2}×\sqrt{\frac{1}{3}}\\&=\frac{\sqrt{3}}{3}\end{aligned}$
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