第126页

信息发布者：
$2\sqrt{3}-2$
解：
$\begin{aligned}&(5+\sqrt{6})\times(5\sqrt{2}-2\sqrt{3})\\=&25\sqrt{2}-10\sqrt{3}+5\sqrt{12}-2\sqrt{18}\\=&25\sqrt{2}-10\sqrt{3}+10\sqrt{3}-6\sqrt{2}\\=&19\sqrt{2}\end{aligned}$
解：
$\begin{aligned}&|1 - \sqrt{3}|-\sqrt{2}\times\sqrt{6}+\frac{1}{2 - \sqrt{3}}-(\frac{2}{3})^{-2}\\=&\sqrt{3}-1 - \sqrt{12}+\frac{2+\sqrt{3}}{(2 - \sqrt{3})(2+\sqrt{3})}-\frac{9}{4}\\=&\sqrt{3}-1 - 2\sqrt{3}+2+\sqrt{3}-\frac{9}{4}\\=&-\frac{5}{4}\end{aligned}$
解：
$\begin{aligned}&(\sqrt{2}+1)^2-\frac{\sqrt{32}\times\sqrt{50}}{\sqrt{8}}\\=&2 + 2\sqrt{2}+1-\frac{\sqrt{1600}}{\sqrt{8}}\\=&3 + 2\sqrt{2}-\frac{40}{2\sqrt{2}}\\=&3 + 2\sqrt{2}-10\sqrt{2}\\=&3 - 8\sqrt{2}\end{aligned}$
解：
$\begin{aligned}&(1 - \sqrt{2})^2-\sqrt{(2\sqrt{2}-3)^2}\\=&1 - 2\sqrt{2}+2-(3 - 2\sqrt{2})\\=&3 - 2\sqrt{2}-3 + 2\sqrt{2}\\=&0\end{aligned}$
解：
$\begin{aligned}&(\sqrt{8}-\sqrt{3})\times\sqrt{8}+(\sqrt{8}-\sqrt{3})\times\sqrt{3}\\=&8-\sqrt{24}+\sqrt{24}-3\\=&5\end{aligned}$

$解:原式=\left[(\sqrt 3+2)(\sqrt3-2)\right]^{2020}×(\sqrt 3-2)-4+4\sqrt 3-3$
$\hspace{1.4cm}=(3-4)^{2020}×(\sqrt 3-2)-7+4\sqrt 3$
$\hspace{1.4cm}=\sqrt 3-2-7+4\sqrt 3$
$\hspace{1.4cm}=5\sqrt{3}-9 $
解：
$\begin{aligned}&|1 - \sqrt{3}|+\sqrt{24}+\frac{1}{2+\sqrt{3}}-(1 - \sqrt{3})^0\\=&\sqrt{3}-1 + 2\sqrt{6}+\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}-1\\=&\sqrt{3}-1 + 2\sqrt{6}+2-\sqrt{3}-1\\=&2\sqrt{6}\end{aligned}$
解：
(1)由题意得$\begin{cases}6 - 2c\geq0\\c - 3\geq0\end{cases}，$即$\begin{cases}c\leq3\\c\geq3\end{cases}，$解得$c = 3。$
$\therefore|a-\sqrt{3}|+\sqrt{b - 4}+\sqrt{6 - 2c}=\sqrt{c - 3}$转化为$|a-\sqrt{3}|+\sqrt{b - 4}=0。$
$\therefore\begin{cases}a-\sqrt{3}=0\\b - 4=0\end{cases}，$解得$a=\sqrt{3},b = 4。$
综上，$a=\sqrt{3},b = 4,c = 3。$
(2)当$a$为腰，$b$为底时，$\sqrt{3},\sqrt{3},4$不能组成三角形，不符合题意；
当$a$为底，$b$为腰时，该三角形的三边长为$4,4,\sqrt{3}，$其周长为$8+\sqrt{3}。$
$\therefore$该等腰三角形的周长为$8+\sqrt{3}。$
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