第18页 - 通城学典课时作业本八年级数学人教版南通专版

$4\sqrt{6}$

$-\frac{2}{3}$

$-a+2b$

$解：原式=(3\sqrt{3})^2-(2\sqrt{2})^2$

$=27-8$

$=19$

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

$=4\sqrt{5}×6\sqrt{2}$

$=24\sqrt{10}$

$解：原式=[(2+\sqrt{5})×(2-\sqrt{5})]^{2023}×(2-\sqrt{5})$

$=(-1)^{2023}×(2-\sqrt{5})$

$=\sqrt{5}-2$

$解：原式=(1+\sqrt{2})^2-(\sqrt{3})^2$

$=1+2\sqrt{2}+2-3$

$=2\sqrt{2}$

$解：∵x+\frac 1x=\sqrt{7} ∴x≠0$

$∴\frac {x^4+x^2+1}{x^2}=x^2+\frac 1{x^2}+1=(x+\frac 1x)^2-1=(\sqrt{7})^2-1=6$

$∴原式=\frac 16$

-2

6

$解：∵m=4-2\sqrt{5}，n=4+2\sqrt{5}$

∴m+n=8，m-n=-4\sqrt_{5}，mn=-4

$∴原式=(m+n)^2-5mn+5(m-n)=64+20-20\sqrt{5}=84-20\sqrt{5}$

$解：∵x=\frac 1{5-2\sqrt{6}}=5+2\sqrt{6}，y=\frac 1{5+2\sqrt{6}}=5-2\sqrt{6}$

$∴x+y=10，xy=1$

$∴原式=\frac {x^2+y^2}{xy}-4=\frac {(x+y)^2-6xy}{xy}=\frac {10^2-6×1}1=94$