第109页 - 苏科版八年级伴你学数学答案（上下册）

$解：ab=(3+2\sqrt{2})(3-2\sqrt{2})=9-8=1$

$a+b=3+2\sqrt{2}+3-2\sqrt{2}=6$

$a^2b+ab^2=ab(a+b)=6$

D

$解:原式=[(2+\sqrt{5})(2-\sqrt{5})]^{2022}(2-\sqrt{5})$

$=2-\sqrt{5}$

$解:原式=[(\sqrt{3}-\sqrt{6})+\sqrt{5}][(\sqrt{3}-\sqrt{6})-\sqrt{5}]$

$=(\sqrt{3}-\sqrt{6})²-(\sqrt{5})²$

$=3-6\sqrt{2}+6-5$

$=4-6\sqrt{2}$

$解：x-y=(2+\sqrt{3})²-(2-\sqrt{3})²=8\sqrt{3}$

$原式=(x-y)²=(8\sqrt{3})²=192$

$解:原式=\sqrt{3+2\sqrt{3}+1}$

$=\sqrt{(\sqrt{3}+1)²}$

$=\sqrt{3}+1$

$解:原式=\sqrt{5-2\sqrt{15}+3}$

$=\sqrt{(\sqrt{5}-\sqrt{3})²}$

$=\sqrt{5}-\sqrt{3}$