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$解:由数轴可知，-2＜a＜-1，2＜b＜3，$

$∴a+1＜0，a+b＞0，b-3＜0. $

$ \begin{aligned}原式&= \sqrt{(a+1)²}+(a+b)+|b-3| \\ &=|a+1|+(a+b)+(3-b) \\ &=(-a-1)+(a+b)+(3-b) \\ &=2. \\ \end{aligned}$

$解:\frac{a²-1}{a+1} -\frac{\sqrt{a²-2a+1}}{a²-a}$

$=\frac{(a+1)(a-1)}{a+1}- \frac{\sqrt{(a-1)²}}{a(a-1)}$

$=(a-1)-\frac {|a-1|}{a(a-1)}.\ $

$∵a=2-\sqrt{3}，\ $

$∴a-1=1-\sqrt{3}＜0.\ $

$ \begin{aligned}∴原式&=a-1+\frac{1}{a}=2-\sqrt {3} -1+2+\sqrt {3} \\ &=3. \\ \end{aligned}$

$ \begin{aligned}解:原式&=3\sqrt{2}÷(\sqrt{3}-2\sqrt{3}) \\ &=3\sqrt{2}÷(-\sqrt{3}) \\ &=- \sqrt{6}. \\ \end{aligned}$

$ \begin{aligned}解:原式&=3 \sqrt{12} ÷2 \sqrt{3} -2 \sqrt{3} ÷2 \sqrt{3}\ \\ &= \frac{3}{2} × \sqrt{\frac{12}{3}} -1 \\ &=3-1 \\ &=2.\ \\ \end{aligned}$

$ \begin{aligned}解:∵ &=\sqrt{5+2\sqrt{6}} = \sqrt{(\sqrt{3})²+2\sqrt{6}+(\sqrt{2})^{2} }\ \\ &= \sqrt{(\sqrt{3}+\sqrt{2})²} = \sqrt{3} + \sqrt{2} ，\ \ \ \ \ \\ y&=\sqrt{5-2\sqrt{6}} = \sqrt{(\sqrt{3})²-2\sqrt{6}+(\sqrt{2})²}\ \\ &= \sqrt{(\sqrt{3}-\sqrt{2})²} = \sqrt{3} - \sqrt{2}，\ \ \ \ \ \\ \end{aligned}$

$∴x+y= \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2} =2 \sqrt{3} ，\ $

$\ xy=( \sqrt{3} + \sqrt{2} )(\sqrt{3} - \sqrt{2} )=3-2=1.\ $