第157页 - 通城学典课时作业本八年级数学苏科版江苏专版

$2\sqrt{2}$

$9-5\sqrt{3}$

$(12-\sqrt{3})$

$3或\sqrt{3}$

$\frac{3\sqrt{2}}{2}或3或6\sqrt{2}-6或6-3\sqrt{2}$

$解：原式=4-\sqrt{6}+2\sqrt{6}$

$=4+\sqrt{6}$

$解：原式=[(3\sqrt2-2\sqrt3)+(3\sqrt2+2\sqrt3)][(3\sqrt2-2\sqrt3)-(3\sqrt2+2\sqrt3)]$

$=6\sqrt2×(-4\sqrt3)$

$=-24\sqrt6$

$解：原式=\sqrt {5}-2-\sqrt {2}×(\frac {\sqrt {2}}4-\frac {\sqrt {10}}2)+\frac 32$

$=\sqrt {5}-2-\frac 12+\sqrt {5}+\frac 32$

$=2\sqrt {5}-1$

$解：原式=2a\sqrt{2a}-2a^2×\frac {\sqrt{2a}}{4a}+3a\sqrt{2a}$

$=2a\sqrt{2a}-\frac a2\sqrt{2a}+3a\sqrt{2a}$

$=\frac 92a\sqrt{2a}$