第158页 - 通城学典课时作业本八年级数学苏科版江苏专版

$ 解：把x=2-\sqrt{3}代入，得$

$原式=(7+4\sqrt{3})×(2-\sqrt{3})^2+(2+\sqrt{3})×(2-\sqrt{3})+\sqrt{3}$

$=(7+4\sqrt{3})×(7-4\sqrt{3})+1+\sqrt{3}$

$=1+1+\sqrt{3}$

$=2+\sqrt{3}$

$ 解：原式=( \frac {1}{x+1} + \frac {1}{x²-1} ) \cdot \frac {x-1}{x}\ $

$= \frac {1}{x+1} \cdot \frac {x-1}{x} +\frac {1}{x²-1} \cdot \frac {x-1}{x}\ $

$= \frac {x-1}{x(x+1)} + \frac {1}{x(x+1)}\ $

$= \frac {x}{x(x+1)}\ $

$= \frac {1}{x+1}$

$当x=\sqrt {12} +( \sqrt {5} )^0-( \frac {1}{2} ) ^{-1}=2 \sqrt {3} +1-2=2 \sqrt {3} -1时，$

$原式=\frac {1}{2\sqrt {3}-1+1} =\frac {\sqrt {3}}{6}$

$解：∵x\sqrt{\frac 2x}+2\sqrt{\frac x2}+\sqrt{18x}=\sqrt{2x}+\sqrt{2x}+3\sqrt{2x}=5\sqrt{2x}=10$

$∴\sqrt{2x}=2$

$解得x=2$

$解：(1)由题意得\begin{cases}8-x≥0\\3x+4≥0\\x+2≥0\end{cases}$

$解得-\frac 43≤x≤8$

$(2)当a^2+b^2=c^2时，8-x+3x+4=x+2$

$解得x=-10(不合题意，舍去)$

$当a^2+c^2=b^2时，8-x+x+2=3x+4$

$解得x=2$

$当b^2+c^2=a^2时，3x+4+x+2=8-x$

$解得x=\frac 25$

$综上所述，x的值为2或\frac 25$