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$解:依题意可知x＜0，y＜0， 所以原式=\sqrt {\frac {x^2}{xy}}+\sqrt {\frac {y^2}{xy}}=\frac {-x}{\sqrt {xy}}+\frac {-y}{\sqrt {xy}}=\frac {-(x+y)}{\sqrt {xy}}.$
$因为x+y=-10，xy=8，所以原式=\frac{-(-10)}{\sqrt{8}}=\frac{5\sqrt 2}{2}.$
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$解:因为a+b=(2+\sqrt{3})+(2-\sqrt{3})=4，$
$ab=(2+\sqrt{3})(2-\sqrt{3})=4-3=1，$
$所以原式=[(a+\sqrt{2})(b+\sqrt{2})]^2=[ab+\sqrt{2}(a+b)+2]^2=(3+4\sqrt{2})^2=41+24\sqrt{2}.$
$解:(1)∵4＜7＜9，∴2＜ \sqrt{7} ＜3.$
$∴b=\sqrt 7-2.$
$∴b(4+b)=( \sqrt{7}-2)(4+\sqrt{7} -2)=( \sqrt{7}-2)( \sqrt{7} +2)=7-4=3.$
$（更多请查看作业精灵详解）$
$解:因为(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1，$
$所以a=\sqrt{3}-\sqrt{2}= \frac{1}{\sqrt 3+\sqrt 2}.$
$同理，b=\frac{1}{2+\sqrt{3}}，c=\frac{1}{\sqrt 5+2}.$
$当分子相同时，分母大的分数反而小，所以a＞b＞c.$
$ \begin{aligned} 解:(1)\frac{1}{2-\sqrt{3}}+\frac{1}{\sqrt 3-\sqrt{2}}&=\frac {2+\sqrt 3}{(2-\sqrt 3)(2+\sqrt 3)}+\frac {\sqrt 3+\sqrt 2}{(\sqrt 3-\sqrt 2)(\sqrt 3+\sqrt 2)} \\ &=2+\sqrt{3} +\sqrt{3} + \sqrt{2} \\ &=2+2 \sqrt{3} + \sqrt{2}. \\ \end{aligned}$
$（更多请查看作业精灵详解）$
$解:∵x=\sqrt 2-1，y=\sqrt{2}+1，$
$∴x+y=2\sqrt{2}，xy=1. $
$ \begin{aligned}∴x^2+y^2&=(x+y)^2-2xy \\ &=(2\sqrt{2})^2-2×1 \\ &=8-2 \\ &=6. \\ \end{aligned}$
$解:∵\sqrt{2022}- \sqrt{2021}=\frac{1}{\sqrt{2022}+\sqrt{2021}}，$
$\sqrt{2021}-\sqrt{2020}=\frac{1}{\sqrt{2021}+\sqrt{2020}}，$
$又∵\sqrt{2022}+ \sqrt{2021}＞ \sqrt{2021}+ \sqrt{2020}， $
$∴\frac{1}{\sqrt{2022}+\sqrt{2021}}＜\frac{1}{\sqrt{2021}+\sqrt{2020}}，$
$∴ \sqrt{2022}- \sqrt{2021}＜\sqrt{2021}- \sqrt{2020}.$
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