第20页 - 通城学典课时作业本八年级数学人教版南通专版

A

C

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18

9

0

$\frac{\sqrt{2}}{2}$

$解：原式=2\sqrt{6}-3\sqrt{2}×\frac {\sqrt{3}}3-18×\frac {\sqrt{6}}6$

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

$=-2\sqrt{6}$

$解：原式=(8\sqrt{3}-3×\frac {\sqrt{6}}3)÷(\sqrt{3})$

$=(8\sqrt{3}-\sqrt{6})÷\sqrt{3}$

$=8-\sqrt{2}$

$解：原式=(6x×\frac {\sqrt {xy}}x+\frac 3{y}×y\sqrt {xy})-(4y×\frac {\sqrt {xy}}y+6\sqrt {xy})$

$=(6\sqrt {xy}+3\sqrt {xy})-(4\sqrt {xy}+6\sqrt {xy})$

$=-\sqrt {xy}$

$当x=\frac 32，y=3时，原式=-\sqrt {\frac 32×3}=-\frac {3\sqrt {2}}2$