第114页 - 江苏版实验班提优训练数学八年级上下册答案

A

D

A

$2\sqrt{3} $

$\sqrt{5} $

-2≤x≤2

-8

解:由题意，得a-1≤0，a+1≥0，

解得-1≤a≤1.

∴原式=2|a-3|+|2a+3|=6-2a+2a+3=9.

$ 解:设CD=x，则4^{2} -x^{2} =6^{2} -(8-x)^{2} ，$

$解得x=\frac {11}{4}.$

$AD=\sqrt {4^{2} -(\frac {11}{4})^{2} }=\frac {3\sqrt {15} }{4}.\ $

$S_{△ABC}=\frac {1}{2}×8 ×\frac {3\sqrt {15} }{4}=3\sqrt {15}.\ $

$\sqrt{5+\frac{1}{7}}=\frac {6}{7}\sqrt {7}\ $

$\ 证明:左边= \sqrt{n+\frac{1}{n+2}} = \sqrt{\frac {n(n+2)+1}{n+2}}\ \ $

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

$= \frac{n+1}{n+2} \sqrt{n+2} =右边\ $

$∴ \sqrt{n+\frac{1}{n+2}} = \frac{n+1}{n+2}\sqrt{n+2}.\ $

$\sqrt{n+\frac{1}{n+2}} = \frac{n+1}{n+2} \sqrt{n+2} $