第159页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

$解:\frac{ab}{a-b}÷(\frac{1}{a+b}+\frac{2b}{a²-b²})$

$=\frac{ab}{a-b}÷\frac{a-b+2b}{(a+b)(a-b)}$

$=\frac{ab}{a-b}·\frac{(a+b)(a-b)}{a+b}$

$=ab$

$当a= \sqrt{5}+1，b=\sqrt{5}-1 时，$

$原式=(\sqrt{5}+1)×(\sqrt{5}-1)=4.$

C

B

C

x＞6

x≤3

$3\sqrt{2}-3$

$ \begin{aligned} 解:原式&=\sqrt{48÷3}-\sqrt{\frac {1}{2}×12}+\sqrt{4×6} \\ &=\sqrt{16}-\sqrt{6}+2\sqrt{6} \\ &=4+\sqrt{6} \\ \end{aligned}$

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&=-4+2(5+1-2\sqrt{5}) \\ &=-4+10+2-4\sqrt{5} \\ &=8-4\sqrt{5} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=(2\sqrt{3}-\frac {1}{5}×5\sqrt{5})-(\sqrt{\frac {1}{5}}-\frac {\sqrt{3}}{9}) \\ &=(2\sqrt{3}-\sqrt{5})-(\frac {\sqrt{5}}{5}-\frac {\sqrt{3}}{9}) \\ &=2\sqrt{3}-\sqrt{5}-\frac {\sqrt{5}}{5}+\frac {\sqrt{3}}{9} \\ &=\frac{19}{9}\sqrt{3}-\frac {6}{5}\sqrt{5} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=(\frac {6\sqrt{x}}{\sqrt{8}}-\sqrt{(2x)^2·\frac {1}{2x}}+\sqrt{3x})÷3\sqrt{2x} \\ &=(\frac {6\sqrt{x}}{2\sqrt{2}}-\sqrt{2x}+\sqrt{3x})·\frac {1}{3\sqrt{2x}} \\ &=(\frac {3}{2}\sqrt{2x}-\sqrt{2x}+\sqrt{3x})·\frac {1}{3\sqrt{2x}} \\ &=(\frac {1}{2}\sqrt{2x}+\sqrt{3x})·\frac {1}{3\sqrt{2x}} \\ &=\frac {1}{2}\sqrt{2x}·\frac {1}{3\sqrt{2x}}+\sqrt{3x}·\frac {1}{3\sqrt{2x}} \\ &=\frac {1}{6}+\frac {1}{3}·\frac {\sqrt{3x}}{\sqrt{2x}} \\ &=\frac {1}{6}+\frac {1}{3}·\sqrt{\frac {3}{2}} \\ &=\frac {1}{6}+\frac {\sqrt{6}}{6} \\ &=\frac {1+\sqrt{6}}{6} \\ \end{aligned}$