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$±\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}\ $

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

$(2)因?yàn)閈frac {\sqrt {2} }{2}÷\sqrt{18}×\sqrt {18} □4\sqrt {2} =-\frac{13}{4}\sqrt {2} ，$

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

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

$因?yàn)閈frac {3\sqrt {2} }{4}-4\sqrt {2}= -\frac{13}{4}\sqrt {2} ，$

$所以□內(nèi)的符號(hào)是“-”.$

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

（更多請(qǐng)點(diǎn)擊查看作業(yè)精靈詳解）

$ \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}$

$ \begin{aligned}解:原式&=\frac {\sqrt {x} \sqrt {y} (\sqrt {x} -\sqrt {y}\ )}{\sqrt {x} \sqrt {y} (\sqrt {x}+\sqrt {y} )}-\frac {\sqrt {x} \sqrt {y} (\sqrt {x} +\sqrt {y} )}{\sqrt {x} \sqrt {y} (\sqrt {y} -\sqrt {x} )} \\ &=\frac {\sqrt {x}- \sqrt {y}}{\sqrt {x}+ \sqrt {y}}-\frac {\sqrt {x} +\sqrt {y}}{\sqrt {y}- \sqrt {x}} \\ &=\frac {(\sqrt {x}- \sqrt {y})^{2} }{(\sqrt {x}+ \sqrt {y})\sqrt {x} -\sqrt {y}}-\frac {(\sqrt {x} +\sqrt {y})^{2} }{(\sqrt {y} -\sqrt {x})(\sqrt {y}+\sqrt {x})} \\ &=\frac {x+y-2\sqrt {xy} }{x-y}-\frac {x+y+2\sqrt {xy} }{y-x} \\ &=\frac {x+y-2\sqrt {xy}+x+y+2\sqrt {xy}\ }{x-y} \\ &=\frac {2(x+y)}{x-y}， \\ \end{aligned}$

$當(dāng)x=3，y=2時(shí)，$

$原式=\frac{2×(3+2)}{3-2}=10.$

$ \begin{aligned} 解:原式&=(\frac {1}{a-\sqrt {ab} }+\frac {1}{\sqrt {ab} +b})·\frac {(\sqrt {a} +\sqrt )(\sqrt {a} -\sqrt )}{\sqrt {ab} } \\ &=[\frac {1}{\sqrt {a} (\sqrt {a} -\sqrt )}+\frac {1}{\sqrt (\sqrt {a}+\sqrt )}]·\frac {(\sqrt {a}+\sqrt )(\sqrt {a} -\sqrt )}{\sqrt {ab} } \\ &=\frac {1}{\sqrt {a} (\sqrt {a} -\sqrt )}·\frac {(\sqrt {a} +-\sqrt )(\sqrt {a} -\sqrt )}{\sqrt {ab} }+\frac {1}{\sqrt (\sqrt {a} +\sqrt )}·\frac {(\sqrt {a}+\sqrt )(\sqrt {a} -\sqrt )}{\sqrt {ab} } \\ &=\frac {\sqrt {a} +\sqrt }{a\sqrt }+\frac {\sqrt {a} -\sqrt }{b\sqrt {a} } \\ &=\frac {\sqrt {ab} +b}{ab}+\frac {a-\sqrt {ab} }{ab} \\ &=\frac {\sqrt {ab}+b+a-\sqrt {ab}\ }{ab} \\ &=\frac {a+b}{ab} \\ \end{aligned}$

$ ∵a=\sqrt {3} +1，b=\sqrt {3} -1， $

$ ∴a+b=\sqrt{3}+1+\sqrt{3}-1=2\sqrt {3} ， $

$ ab=(\sqrt {3} +1)(\sqrt {3} -1)=2， $

$ 則\frac{a+b}{ab}=\frac{2\sqrt {3} }{2}=\sqrt{3}. $

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