第44页 - 启东中学作业本八年级数学苏科版宿迁专版

$\frac {\sqrt {ab}}{b}$

$\sqrt b$

能开得尽方

被开方数

分母

D

C

$ \begin{aligned}解:原式&=\frac {1}{3\sqrt 2} \\ &=\frac{\sqrt{2}}{6} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {\sqrt 8}{8} \\ &=\frac{\sqrt{2}}{4} \\ \end{aligned}$

$ 解:原式=\sqrt {\frac {16}{5}} $

$\hspace{1.4cm}=\frac {4}{\sqrt 5} $

$\hspace{1.4cm}=\frac{4\sqrt{5}}{5} $

$ \begin{aligned} 解:原式&=\frac {\sqrt 2}{\sqrt 5} \\ &=\frac {\sqrt {10}}{5} \\ \end{aligned}$

$解:原式=\frac{\sqrt{3}}{3}$

$ \begin{aligned}解:原式&=\frac {\sqrt 2}{2\sqrt 2} \\ &=\frac{1}{2} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {\sqrt {3y}}{2x\sqrt {2x}} \\ &=\frac {\sqrt {6xy}}{4x^2} \\ \end{aligned}$

$解:∵a=\sqrt 5+2，b=\sqrt 5-2，\ $

$∴a^2=(\sqrt{5}+2)^2=9+4\sqrt 5，b^2=(\sqrt{5}-2)^2=9-4\sqrt{5}，\ $

$∴原式= \sqrt{9+4\sqrt{5}+9-4\sqrt{5}+7}= \sqrt{25}=5.$

$ab$