第121页 - 通城学典课时作业本八年级数学人教版南通专版

$\frac{4\sqrt{3}}{3}$

-1

4

$20\sqrt{3}$

$-\sqrt{3}+4\sqrt{2}$

$解：原式=3\sqrt{6}-\frac {\sqrt{6}}3-2×\frac{\sqrt{2}}2+4\sqrt{2}$

$=\frac 83\sqrt{6}+3\sqrt{2}$

$解：原式=-3×\frac {2\sqrt{6}}9÷\frac {\sqrt{3}}2×3\sqrt{3}$

$=-4\sqrt{6}$

$解：原式=(-2\sqrt{6})^2-(3\sqrt{2})^2$

$=24-18$

$=6$

$解：原式=[(2\sqrt{7}+5\sqrt{2})+(2\sqrt{7}-5\sqrt{2})][(2\sqrt{7}+5\sqrt{2})-(2\sqrt{7}-5\sqrt{2})]$

$=4\sqrt{7}×10\sqrt{2}$

$=40\sqrt{14}$

$解：原式=\frac 12×2\sqrt{3}-3×\frac {\sqrt{3}}3-\sqrt{2}$

$=\sqrt{3}-\sqrt{3}-\sqrt{2}$

$=-\sqrt{2}$

$解：原式=\frac {\sqrt{6}}{\sqrt{3}+\sqrt{2}}×\frac {\sqrt{3}×\sqrt{2}}{\sqrt{2}-\sqrt{3}}$

$=\frac {\sqrt{6}×\sqrt{6}}{(\sqrt{3}+\sqrt{2})(\sqrt{2}-\sqrt{3})}$

$=-6$

$解：∵x=\sqrt{3}+\sqrt{7}，y=\sqrt{3}-\sqrt{7}$

$∴x-y=2\sqrt{7}，xy=-4$

$∴原式=4(x-y)^2+xy=4×(2\sqrt{7})^2+(-4)=108$