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解：​$(3)$​原式​$=5-2\sqrt {10}+2-2\sqrt {10}$​
​$=7-4\sqrt {10}$​
解：​$(4)$​原式​$=\sqrt {3}×\sqrt {8}-\frac {1}{2}×3+4\sqrt {2}×\sqrt {8}-2\sqrt {2}×\sqrt {3}$​
​$=\sqrt {24}-\frac {3}{2}+4\sqrt {16}-2\sqrt {6}$​
​$=2\sqrt {6}-\frac {3}{2}+16-2\sqrt {6}$​
​$=14\frac {1}{2}$​
解：由数轴可知​$b＜a＜0＜c$​，且​$|b|＞|c|＞|a|$​，
​$ $​则​$\sqrt {a^2}=-a$​，​$|a+b|=-(a+b)$​，​$\sqrt {(c-a)^2}=c-a$​，​$|b+c|=-(b+c)$​，
​$ $​原式​$=-a-(-a-b)+c-a+(-b-c)$​
​$ =-a+a+b+c-a-b-c$​
​$ =-a$​
解：
$\begin{aligned}(1-\frac{2}{a+1})÷\frac{a^2-2a+1}{a+1}&=\frac{a+1-2}{a+1}×\frac{a+1}{(a-1)^2}\\&=\frac{a-1}{a+1}×\frac{a+1}{(a-1)^2}\\&=\frac{1}{a-1}\end{aligned}$
当$a=\sqrt{2}+1$时，
原式$=\frac{1}{\sqrt{2}+1-1}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
解：​$x=\frac {1}{2-\sqrt {3}}=\frac {2+\sqrt {3}}{(2-\sqrt {3})(2+\sqrt {3})}=2+\sqrt {3}$​，则​$\frac {1}{x}=2-\sqrt {3}$​
​$ (1) x+\frac {1}{x}=2+\sqrt {3}+2-\sqrt {3}=4$​
​$ (2)$​
​$ \begin {aligned}(7-4\sqrt {3})x^2+(2-\sqrt {3})x+\sqrt {3}&=(7-4\sqrt {3})(2+\sqrt {3})^2+(2-\sqrt {3})(2+\sqrt {3})+\sqrt {3}\\&=(7-4\sqrt {3})(7+4\sqrt {3})+(4-3)+\sqrt {3}\\&=49-48+1+\sqrt {3}\\&=2+\sqrt {3}\end {aligned}$​
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