第141页 - 经纶学典学霸八年级数学上下册【江苏国标】

$±\sqrt{7} $

2

$ \begin{aligned}解:原式&=2+ \sqrt{5}-2-9+3-\sqrt{5} \\ &=-6. \\ \end{aligned}$

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

$解:(1)\frac{\sqrt {2} }{2}+\ \sqrt{8} - \sqrt{18} -4\sqrt {2}\ $

$= \frac{\sqrt {2} }{2} +2\sqrt {2} -3\sqrt {2} -4\sqrt {2} = -\frac{9\sqrt {2} }{2}.$

$(2)因为\frac {\sqrt {2} }{2}÷\sqrt{18}×\sqrt {18} □4\sqrt {2} =-\frac{13}{4}\sqrt {2} ，$

$所以\frac{\sqrt {2} }{2}×\frac{1}{2\sqrt {2} }×3\sqrt {2}□ 4\sqrt {2} =-\frac{13}{4}\sqrt {2} ，$

$所以\frac{3\sqrt {2} }{4}□4\sqrt {2} =-\frac{13}{4}\sqrt {2} .$

$因为\frac {3\sqrt {2} }{4}-4\sqrt {2}= -\frac{13}{4}\sqrt {2} ，$

$所以□内的符号是“-”.$

$(3)12-\frac{7\sqrt {2} }{2}$

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&= (3\sqrt {3} ×3\sqrt {6} + \frac{4}{5} ×5\sqrt {2} -8× \frac{\sqrt {2} }{2} )÷\sqrt {2} \\ &=(27\sqrt {2} +4\sqrt {2} - 4\sqrt {2} ) ÷\sqrt{2} \\ &=27\sqrt {2} ÷\sqrt {2} \\ &=27. \\ \end{aligned}$

$ \begin{aligned}解:原式&=[\sqrt {2} +(\sqrt {3} - \sqrt{5} )]×[\sqrt {2} -(\sqrt {3} -\sqrt {5} )] \\ &=(\sqrt {2} )²-(\sqrt {3} - \sqrt{5} )² \\ &=2-(3-2 \sqrt{15} +5) \\ &=2-8+2\sqrt {15} \\ &=-6+2 \sqrt{15}. \\ \end{aligned}$