第105页 - 苏科版八年级伴你学数学答案（上下册）

$解：原式=\frac {\sqrt{2y}}{3x\sqrt{2x}}$

$=\frac {\sqrt{xy}}{3x^2}$

$解：原式=-\sqrt{(a-1)^2×(-\frac 1{a-1})}$

$\ \ \ \ \ \ \ \ =-\sqrt{1-a}$

$解:原式=\frac {\sqrt{11}×\sqrt{3}}{\sqrt{3}×\sqrt{3}}$

$=\frac {\sqrt{33}}{3}$

$解:原式=\frac {\sqrt{14}×\sqrt{6}}{\sqrt{6}×\sqrt{6}}$

$=\frac {\sqrt{21}}{3}$

$解:原式=\frac {3\sqrt{y}}{x\sqrt{y}×\sqrt{y}}$

$=\frac {3\sqrt{y}}{xy}$

$解：(1)\frac {1}{1+\sqrt{2}}=\frac {1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}=-(1-\sqrt{2})=-1+\sqrt{2}$

$(2)\frac 1{\sqrt{n}+\sqrt{n+1}}=-\sqrt{n}+\sqrt{n+1}$

$(3)原式=(-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}+...-\sqrt{2014}+\sqrt{2015})×(1+\sqrt{2015})$

$ =(-1+\sqrt{2015})(1+\sqrt{2015})$

$ =2015-1$

$ =2014$