第112页 - 苏科版八年级伴你学数学答案（上下册）

D

-1

$\sqrt{5}$或$\sqrt{13}$

解:原式$=\frac {a+2-3}{(a+2)(a-2)}×\frac {a(a+2)}{a-1}$

$=\frac {a}{a-2}$

当$a=\sqrt{6}+2$时

原式$=\frac {\sqrt{6}+2}{\sqrt{6}+2-2}$

$=\frac {3+\sqrt{6}}{3}$

解：$(1)$猜想，$\sqrt{2011}-\sqrt{2010}＞\sqrt{2012}-\sqrt{2011}$

$\sqrt{2011}-\sqrt{2010}=\frac {(\sqrt{2011}-\sqrt{2010})(\sqrt{2011}+\sqrt{2010})}{\sqrt{2011}+\sqrt{2010}}=\frac 1{\sqrt{2011}+\sqrt{2010}}$

$\sqrt{2012}-\sqrt{2011}=\frac {(\sqrt{2012}-\sqrt{2011})(\sqrt{2012}+\sqrt{2011})}{\sqrt{2012}+\sqrt{2011}}=\frac 1{\sqrt{2012}+\sqrt{2011}}$

$∵\sqrt{2011}+\sqrt{2010}＜\sqrt{2012}+\sqrt{2011}$

$∴\frac 1{\sqrt{2011}+\sqrt{2010}}＞\frac 1{\sqrt{2012}+\sqrt{2011}}，$即$\sqrt{2011}-\sqrt{2010}＞\sqrt{2012}-\sqrt{2011}$

$(2)$由$(1)$可得

原式$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+....+10-\sqrt{99}$

$=10-1$

$=9$