第137页 - 经纶学典学霸八年级数学上下册【江苏国标】

D

$x=\frac{\sqrt {30}}{2}$

5

$解:(1)方法一:\frac{2}{\sqrt {5}+\sqrt{3}}=\frac{2×(\sqrt{5}-\sqrt {3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\sqrt{5}-\sqrt{3}$

$方法二:\frac{2}{\sqrt {5}+\sqrt{3}}=\frac{(\sqrt{5})^{2}-(\sqrt {3})^{2}}{\sqrt{5}+\sqrt{3}}=\sqrt{5}-\sqrt{3}$

$(2)\frac{3}{\sqrt{11}-2\sqrt {2}}=\frac{3×(\sqrt{11}+2\sqrt {2})}{(\sqrt{11}-2\sqrt {2})(\sqrt{11}+2\sqrt {2})}=\sqrt{11}+2\sqrt {2}$

$\frac{4}{\sqrt{15}-\sqrt{11}}=\frac{4×(\sqrt{15}+\sqrt{11})}{(\sqrt{15}-\sqrt{11})(\sqrt{15}+\sqrt{11})}=\sqrt{15}+ \sqrt{11}$

$∵2\sqrt {2}＜ \sqrt{15}∴ \sqrt{11}+2\sqrt {2}＜ \sqrt{15}+ \sqrt{11}$

$∴\frac{3}{\sqrt{11}-2\sqrt {2}}＜\frac{4}{\sqrt{15}-\sqrt {11}}$

3

$解:(1)\sqrt{15}-\sqrt{14}=\frac{1}{\sqrt{15}+\sqrt{14}}$

$\ \sqrt{14}-\sqrt{13}=\frac{1}{\sqrt{14}+\sqrt {13}}$

$∵ \sqrt{15}＞ \sqrt {13}，∴ \sqrt{15}+\sqrt{14}＞ \sqrt{14}+ \sqrt{13}$

$∴ \sqrt{15}- \sqrt{14}＜ \sqrt{14}- \sqrt{13}$

$(2)∵x+1≥0,x-1≥0,∴x≥1$

$∵y= \sqrt{x+1}- \sqrt{x-1}+3=\frac {2}{\sqrt {x+1}+\sqrt {x-1}}+3$

$当x=1时，分母 \sqrt{x+1}+\sqrt{x-1}有最小值\sqrt {2}$

$∴y=\frac {2}{\sqrt {x+1}+\sqrt {x-1}}+3的最大值是\sqrt {2}+3$