第17页 - 通城学典课时作业本八年级数学人教版南通专版

$解：原式=(3\sqrt{3}+3\sqrt{5})×(2\sqrt{3}-2\sqrt{5})-(2\sqrt{3}+3\sqrt{2})^2$

$=18-6\sqrt{15}+6\sqrt{15}-30-(12+12\sqrt{6}+18)$

$=18-30-12-12\sqrt{6}-18$

$=-42-12\sqrt{6}$

$解：∵x=\sqrt{3}-2，y=\sqrt{3}+2$

$∴x+y=2\sqrt{3}，xy=-1$

$∴原式=3(x+y)^2-8xy=3×(2\sqrt{3})^2-8×(-1)=44$

$解：∵6+4\sqrt{2}=(2+\sqrt{2})^2$

$∴原式=(2+\sqrt{2})^2x^2+(2+\sqrt{2})x+\sqrt{2}=[(2+\sqrt{2})x]^2+(2+\sqrt{2})x+\sqrt{2}$

$当x=2-\sqrt{2}时，原式=[(2+\sqrt{2})×(2-\sqrt{2})]^2+(2+\sqrt{2})×(2-\sqrt{2})+\sqrt{2}=4+2+\sqrt{2}=6+\sqrt{2}$

$解:原式=\frac{x+1+2}{x+1}×\frac{x+1}{(x-3)(x+3)}$

$=\frac{x+3}{x+1}×\frac{x+1}{(x+3)(x-3)}$

$=\frac{1}{x-3}$

$当x=\sqrt{3}+3时，原式=\frac{\sqrt{3}}{3}$

$解：x=\frac 1{\sqrt{5}-\sqrt{3}}=\frac {\sqrt{5}+\sqrt{3}}2，y=\frac 1{\sqrt{5}+\sqrt{3}}=\frac {\sqrt{5}-\sqrt{3}}2$

$∴x+y=\sqrt{5}，xy=\frac 12$

$(1)原式=(x+y)^2-3xy=5-\frac 32=\frac 72$

$(2)原式=\frac {x^2+y^2}{xy}=\frac {(x+y)^2-2xy}{xy}=\frac {5-1}{\frac 12}=8$