第143页 - 经纶学典学霸八年级数学上下册【江苏国标】

D

$2\sqrt {3}\ $

$-\sqrt {6}\ $

$\frac{\sqrt {\ a-b}}{a-b}$

C

$3\sqrt {5}\ $

$\frac{2\sqrt {2} +\sqrt {3} }{5}$

$解:∵\frac{1}{3-\sqrt {7} }=\frac{3+7}{(3+\sqrt{7})(3-\sqrt {7} )} =\frac{3+\sqrt {7} }{2}，$

$又2＜\sqrt {7}＜ 3，∴\frac{5}{2}＜\frac{3+\sqrt {7} }{2}＜3，\ $

$∴a=2，b=\frac{3+\sqrt{7}}{2}-2=\frac{\sqrt {7} -1}{2}，$

$∴\frac{a}{b}=2÷\frac {\sqrt {7} -1}{2}=\frac{4}{\sqrt {7} -1}=\frac{2\sqrt {7} +2}{3}.$

$ \begin{aligned}解:原式&=\frac{(2+\sqrt {3} )^{2} }{2+\sqrt {3} } \\ &=2+\sqrt {3} . \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt {y} } \\ &=\sqrt{x} +\sqrt{y} \\ \end{aligned}$

（更多请点击查看作业精灵详解）

$解:∵ x＞2，$

$∴ \frac{x-2+\sqrt{x²-4}}{x+2+\sqrt{x²-4}}\ $

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

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

$=\frac{\sqrt {x-2}}{\sqrt{x+2}}$

$=\frac{\sqrt {x²-4}}{x+2}.$

$解:\frac{\sqrt {2} +2\sqrt {3}\ +\sqrt {5}\ }{(\sqrt {2}\ +\sqrt {3}\ )(\sqrt {3}\ +\sqrt {5}\ )}\ $

$=\frac{(\sqrt {2}\ +\sqrt {3}\ )+(\sqrt {3}\ +\sqrt{5})}{(\sqrt {2}\ +\sqrt {3}\ )(\sqrt {3} +\sqrt{5})}\ $

$=\frac {\sqrt {2} +\sqrt {3}\ }{(\sqrt {2}+\sqrt {3}\ )(\sqrt {3} +\sqrt {5}\ )}+\frac {\sqrt {3} +\sqrt {5} }{(\sqrt {2}+\sqrt {3}\ )(\sqrt {3} +\sqrt {5} )}$

$=\frac {1}{\sqrt {3}+\sqrt {5}}+\frac {1}{\sqrt {2}+\sqrt {3}}$

$=\frac {\sqrt {5}+\sqrt {3}-2\sqrt {2} }{2}.$

$原式=\frac {(\sqrt {2}+\sqrt {5}-\sqrt {3} )^{2} }{(\sqrt {2}+\sqrt {5}+\sqrt {3}\ )(\sqrt {2}+\sqrt {5}-\sqrt {3}\ )}$

$=\frac {(\sqrt {2}+\sqrt {5}-\sqrt {3}\ )^{2} }{(\sqrt {2}+\sqrt {5})^{2} -3}$

$=\frac {\sqrt {10}-\sqrt {6}-\sqrt {15}+5 }{\sqrt {10} +2}$

$=\frac {(\sqrt {10}-\sqrt {6}-\sqrt {15}+5)(\sqrt {10} -2)}{(\sqrt {10} +2)(\sqrt {10} -2)}$

$=\frac {3\sqrt {10} -3\sqrt {6} }{6}$

$=\frac {\sqrt {10} -\sqrt {6} }{2}.$