第143页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

2

$\frac {5\sqrt{2}}{2}-2\sqrt{3}$

$5\sqrt{7}-13$

$解:∵a=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}＜1， $

$ \begin{aligned}∴原式&=\frac{(a-1)²}{a-1}-\frac{\sqrt{(a-1)²}}{a(a-1)} \\ &=a-1+\frac{1}{a} \\ &=2-\sqrt{3}-1+2+\sqrt{3} \\ &=3. \\ \end{aligned}$

（更多请点击查看作业精灵详解）

$=\frac 34×2\sqrt 6÷(-\sqrt 3)×\frac 14×3\sqrt 6$

$=-\frac {9\sqrt 3}4$

$=3\sqrt 3+3×\frac {2\sqrt 3}3-\frac {4\sqrt 3}3-\frac 23×4\sqrt 3$

$=3\sqrt 3+2\sqrt 3-\frac {4\sqrt 3}3-\frac {8\sqrt 3}3$

$=\sqrt 3$

$=3-2\sqrt 3+1-(6\sqrt 3-12+9-6\sqrt 3)-2\sqrt {\frac 23×6}$

$=4-2\sqrt 3+3-4$

$=3-2\sqrt 3$

$=[\sqrt 5-(\sqrt 3-\sqrt 2)][\sqrt 5+(\sqrt 3-\sqrt 2)]$

$=5-(\sqrt 3-\sqrt 2)^2$

$=5-(3-2\sqrt 6+2)$

$=2\sqrt 6$

$解:由题意得\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

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

$=5-2\sqrt{6}$

$设x=\sqrt{6-3\sqrt{3}}- \sqrt{6+3\sqrt{3}}，$

$易知 \sqrt{6-3\sqrt{3}}＜ \sqrt{6+3\sqrt{3}}，$

$故x＜0，$

$ \begin{aligned} 由x²&=(\sqrt{6-3\sqrt{3}} -\sqrt{6+3\sqrt{3}}~~~)² \\ &=6-3\sqrt{3}+6+3\sqrt{3}-2 \sqrt{(6-3\sqrt{3})(6+3\sqrt{3}}) \\ &=6， \\ \end{aligned}$

$解得x=- \sqrt{6}，$

$即 \sqrt{6-3\sqrt{3}}-\sqrt{6+3\sqrt{3}}=-\sqrt{6}，$

$则\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}+ \sqrt{6-3\sqrt{3}}- \sqrt{6+3\sqrt{3}}$

$=5-2\sqrt{6}-\sqrt{6}$

$=5-3\sqrt{6}$