第126页 - 江苏版实验班提优训练数学八年级上下册答案

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

$当a=4+\sqrt {3} ，b=4- \sqrt{3}时，\ $

$原式=\frac{4+\sqrt {3} +4-\sqrt{3}}{(4+\sqrt {3} )-(4-\sqrt{3})}=\frac{8}{2\sqrt{3}}=\frac{4\sqrt{3}}{3}.$

$解:设a+2\sqrt{6}=m，则a=m-2\sqrt{6}，$

$设\frac{1}{a}-2\sqrt{6}=n，则a=\frac{1}{n+2\sqrt{6}}.\ $

$∴m-2\sqrt{6}=\frac{1}{n+2\sqrt{6}}.\ $

$∴(m-2\sqrt{6})(n+2\sqrt{6})=1.\ $

$∴mm-2\sqrt{6} n+2\sqrt{6} m=25.\ $

$∵m、n为整数，∴m=n=5或m=n=-5.\ $

$∴a=5-2\sqrt{6}或a=-5-2\sqrt{6} . $