第141页 - 经纶学典学霸八年级数学上下册【江苏国标】

$3\sqrt {5}+3$

$解:原式=2+\sqrt{5}-2-9+3-\sqrt {5}=-6$

$解:原式=\sqrt {2}×(\sqrt {2}+\frac {\sqrt {2}}{2})-\frac{3\sqrt {2}-2\sqrt {2}}{\sqrt {2}}$

$=3-1$

$=2$

$解:原式=(27\sqrt {2}+4\sqrt {2}- 4\sqrt {2})÷\sqrt {2}$

$=27$

$解:原式=[\sqrt {2}+(\sqrt {3}-\sqrt{5})]×[\sqrt {2}-(\sqrt {3}-\sqrt{5})]$

$=2-(3-2 \sqrt{15}+5)$

$=2-8+2\sqrt{15}$

$=-6+2 \sqrt{15}$

A

$解:原式=\frac{\sqrt{x}\sqrt{y}(\sqrt{x}-\sqrt{y})}{\sqrt{x}\sqrt{y}(\sqrt{x}+\sqrt{y})}-\frac{\sqrt{x}\sqrt{y}(\sqrt{x}+\sqrt{y})}{\sqrt{x}\sqrt{y}(\sqrt{y}-\sqrt{x})}$

$=\frac {(\sqrt {x}-\sqrt {y})^{2}}{(\sqrt {x}+\sqrt {y})(\sqrt {x}-\sqrt {y})}-\frac {(\sqrt {x}+\sqrt {y})^{2}}{(\sqrt {y}-\sqrt {x})(\sqrt {y}+\sqrt {x})}$

$=\frac {x+y-\sqrt {xy}+x+y+2\sqrt {xy}}{x-y}=\frac {2(x+y}{x-y}$

$把x=3,y=2代入原式=10$

$解:原式=[\frac {1}{\sqrt {a}(\sqrt {a}-\sqrt {b})}+\frac {1}{\sqrt {b}(\sqrt {a}+\sqrt {b})}]×\frac {(\sqrt {a}+\sqrt {b})(\sqrt {a}-\sqrt {b})}{\sqrt {ab}}$

$=\frac {\sqrt {a}+\sqrt {b}}{a\sqrt {b}}+\frac {\sqrt {a}-\sqrt {b}}{b\sqrt {a}}$

$=\frac {a+b}{ab}$

$把a,b的值代入原式=\frac {2\sqrt {3}}{2}=\sqrt {3}$

$解:(1)原式=\frac {\sqrt {2}}{2}+2\sqrt {2}-3\sqrt {2}-4\sqrt {2}=-\frac{9\sqrt {2}}{2}$

$(2)由题\frac{\sqrt {2}}{2}×\frac{1}{2\sqrt {2}}×3\sqrt {2}口 4\sqrt {2}=-\frac{13}{4}\sqrt {2}$

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

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

$所以口内的符号是“-”$

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