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第21页

信息发布者：

 解：$(\sqrt{12}+\sqrt{8})×\sqrt{3}-\sqrt{6}$

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

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

 $=6+\sqrt{6}$

 解：$\sqrt{27}÷3-\sqrt{\frac{3}{2}}×\sqrt{12}+\sqrt{72}$

 $=3\sqrt{3}÷3-\sqrt{18}+6\sqrt{2}$

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

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

 解：$(\sqrt{2}-1)^2+\sqrt{32}-(\sqrt{5}+3)(\sqrt{5}-3)$

 $=2-2\sqrt{2}+1+4\sqrt{2}-(5-9)$

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

 $=7+2\sqrt{2}$

 解：$(\sqrt{3}+1)(\sqrt{3}-1)-(\sqrt{18}-\sqrt{24})÷\sqrt{6}$

 $=3-1-(\sqrt{3}-2)$

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

 $=4-\sqrt{3}$

 解：

 (1) $\because x=1-\sqrt{2},y=1+\sqrt{2}$

 $\therefore x+y=2，$$xy=(1-\sqrt{2})(1+\sqrt{2})=1-2=-1$

 $\therefore \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{-1}=-2$

 (2) $x^2+y^2+xy+2x-2y$

 $=(x+y)^2-xy+2(x-y)$

 $\because x-y=(1-\sqrt{2})-(1+\sqrt{2})=-2\sqrt{2}$

 $\therefore$ 原式$=2^2-(-1)+2×(-2\sqrt{2})$

 $=4+1-4\sqrt{2}$

 $=5-4\sqrt{2}$

$\frac{3-\sqrt{6}}{3}$

 解：$\because a=\frac{1}{3+\sqrt{10}}=\frac{3-\sqrt{10}}{(3+\sqrt{10})(3-\sqrt{10})}=\sqrt{10}-3$

 $\therefore a+3=\sqrt{10}$

 $\therefore a^2+6a-27=(a+3)^2-36$

 $=(\sqrt{10})^2-36$

 $=10-36$

 $=-26$