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$\sqrt{2}-1\ $
$ \sqrt{7}-\sqrt{6}$
$ 解:原式=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}$
$-\sqrt{3}+···+ \sqrt{10}-\sqrt{9} $
$= \sqrt{10}-1. $
（更多请点击查看作业精灵详解）
（更多请点击查看作业精
灵详解）
$解:方法一:$
$\frac{3}{\sqrt {6} -\sqrt{3}}\ $
$=\frac {3×(\sqrt{6}+\sqrt{3})\ }{ (\sqrt{6}-\sqrt{3})×( \sqrt{6}+\sqrt{3})}\ $
$=\frac{3×(\sqrt{6}+\sqrt{3})}{6-3}$
$=\sqrt{6}+\sqrt{3}；\ $
$方法二:$
$\frac{3}{\sqrt {6} -\sqrt{3}}$
$=\frac {6-3}{\sqrt{6}-\sqrt{3}\ }$
$=\frac{(\sqrt{6})²-(\sqrt{3})^{2} }{\sqrt{6}-\sqrt{3}}\ $
$=\frac {(\sqrt {6} +\sqrt {3} )×(\sqrt {6} -\sqrt {3} )}{\sqrt {6} -\sqrt {3} }$
$=\sqrt{6}+\sqrt{3}. $
$ \begin{aligned}解:原式&=\frac{\sqrt{5}(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \\ &=\frac {5+\sqrt {5} }{5-1} \\ &= \frac {5+\sqrt {5} }{4}. \\ \end{aligned}$
$解:原式=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}+\frac {\sqrt {3}-\sqrt {2}\ }{(\sqrt {3}+\sqrt {2}\ )(\sqrt {3}-\sqrt {2}\ )}$
$+\frac {\sqrt {4} -\sqrt {3} }{(\sqrt {4} +\sqrt {3} )(\sqrt {4} -\sqrt {3} )}+···+\frac {\sqrt{2023}-\sqrt{2022} }{(\sqrt{2023}-\sqrt{2022}) (\sqrt{2023}+\sqrt {2022} )}$
$= \sqrt{2}-1+ \sqrt{3}- \sqrt{2}+ \sqrt{4}- \sqrt{3}+···+\sqrt{2023}- \sqrt{2022}$
$= \sqrt{2023}-1.$
$解:\sqrt{8}-\sqrt{7}＜\sqrt{6}-\sqrt{5}，理由如下:\ $
$\sqrt {8} -\sqrt {7} =\frac { \sqrt{8}-\sqrt{7}}{8-7}$
$= \frac {\sqrt {8}-\sqrt {7}\ }{(\sqrt {8} +\sqrt {7} )(\sqrt {8} -\sqrt {7} )}=\frac {1}{\sqrt {8} +\sqrt {7} }，$
$\sqrt{6}-\sqrt{5}=\frac {\sqrt {6} -\sqrt {5} }{6-5}$
$=\frac {\sqrt {6} -\sqrt {5} }{(\sqrt {6} +\sqrt {5} )(\sqrt {6} -\sqrt {5} )}$
$=\frac {1}{\sqrt {6}+\sqrt {5}\ }.$
$∵\sqrt{8}＞\sqrt {7} ＞ \sqrt{6}＞\sqrt {5} ，\ $
$∴\sqrt{8}+\sqrt{7}＞ \sqrt{6}+\sqrt{5}，\ $
$∴\frac{1}{\sqrt {8} +\sqrt{7}}＜\frac{1}{\sqrt {6} +\sqrt{5}}，$
$∴\sqrt{8}-\sqrt{7}＜\sqrt{6}-\sqrt{5}.$
$ \begin{aligned}解:原式&=\frac {\sqrt {3}-\sqrt {2} -1 }{(\sqrt {3} +\sqrt {2}+1 )(\sqrt {3} -\sqrt {2} -1)} \\ &=\frac {\sqrt {3} -\sqrt {2} -1}{(\sqrt {3} )^{2} -(\sqrt {2} +1)^{2} } \\ &=\frac {\sqrt {3}-\sqrt {2}-1\ }{3-(3+2\sqrt {2} )} \\ &=\frac {\sqrt {3} -\sqrt {2} -1}{12\sqrt {2} } \\ &=\frac {-\sqrt {2}(\sqrt {3} -\sqrt {2-1} ) }{-2\sqrt {2} ×(-\sqrt {2} )} \\ &=\frac {2+\sqrt {2} -\sqrt {6} }{4}. \\ \end{aligned}$
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