第113页 - 苏科版数学补充习题八年级上下册答案

$解：原式 =2\sqrt {2}-{\sqrt {2}×{\sqrt {2}-{\sqrt {2}}}}×2$

$=2\sqrt {2}-2-2\sqrt {2}$

$=-2$

$解：原式 ={\frac {2\sqrt {5}+{\sqrt {5}}} {\sqrt {5}}}-{\sqrt {\frac {1} {3}×12}}$

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

$=3-2$

$=1$

$解：原式 ={\frac {(x+y)-(x-y)} {(x+y)(x-y)}}÷{\frac {2y} {{(x+y)}^{2}}}$

$={\frac {2y} {(x+y)(x-y)}}·{\frac {{(x+y)}^{2}} {2y}}$

$={\frac {x+y} {x-y}}$

$将x=\sqrt {3}+\sqrt {2}，y=\sqrt {3}-\sqrt {2}代入，得$

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

$={\frac {2\sqrt {3}} {2\sqrt {2}}}$

$={\frac {\sqrt {6}} {2}}$

$解：由题意得，$

${a}^{2}=96×12$

$∵ a＞0$

$∴ a={\sqrt {96×12}}=24{\sqrt {2}}(cm)$

$答：a的值为24\sqrt {2}cm。$

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