第131页 - 通城学典课时作业本八年级数学苏科版江苏专版

5

$\frac{1}{3}$

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

$2-\sqrt{3}或-2+\sqrt{3}$

$ \begin{aligned}解：原式&=4\sqrt3×(12\sqrt3-\frac 43×\frac {\sqrt2}4-3×3\sqrt3) \\ &=4\sqrt3×(12\sqrt3-\frac {\sqrt2}3-9\sqrt3) \\ &=144-\frac 43\sqrt6-108 \\ &=36-\frac 43\sqrt6 \\ \end{aligned}$

$ \begin{aligned}解：原式&=(20\sqrt3+2\sqrt3-7\sqrt7)÷\sqrt3 \\ &=20+2-7\frac {\sqrt7}{\sqrt3} \\ &=22-\frac 73\sqrt{21} \\ \end{aligned}$

$ \begin{aligned}解：原式&=\frac {(\sqrt 5-2)^2}{9} \\ &=\frac {5-4\sqrt 5+4}9 \\ &=\frac {9-4\sqrt 5}9 \\ \end{aligned}$

$ \begin{aligned}解：原式&=(2\sqrt{5})^2-(5\sqrt{2})^2-(5+2-2\sqrt{10}) \\ &=20-50-5-2+2\sqrt{10} \\ &=-37+2\sqrt{10} \\ \end{aligned}$

$ \begin{aligned} 解：原式&=\frac x{x-4} \cdot \frac {(x+2)(x-2)-x(x-1)}{x(x-2)^2} \\ &=\frac x{x-4} \cdot \frac {x-4}{x(x-2)^2} \\ &=\frac 1{(x-2)^2} \\ \end{aligned}$

$ 当x=3-\sqrt {2}时，原式=\frac 1{(3-\sqrt {2}-2)^2}=\frac 1{(1-\sqrt {2})^2}=\frac 1{3-2\sqrt {2}}=3+2\sqrt {2} $

$解：若“□”填-\sqrt {2}，“○”填\sqrt {3}$

$则(-\sqrt {2}+\sqrt {3})^2÷\sqrt {2}=(5-2\sqrt {6})÷\sqrt {2}=\frac {5\sqrt {2}}2-2\sqrt {3}$