第13页 - 通城学典课时作业本八年级数学人教版南通专版

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$解:(1)原式=3\sqrt{6}+\frac {\sqrt{2}}2-\frac {\sqrt{2}}4+\sqrt{6}$

$=4\sqrt{6}+\frac {\sqrt{2}}4$

$解:(2)原式=10\sqrt{2x}-5\sqrt{2x}-6×\frac {\sqrt{2x}}2+\frac {3\sqrt{2x}}{2}$

$=10\sqrt{2x}-5\sqrt{2x}-3\sqrt{2x}+\frac {3\sqrt{2x}}2$

$=\frac {7\sqrt{2x}}2$

$解：x^2+4y^2-6x-4y+10=x^2-6x+9+4y^2-4y+1=(x-3)^2+(2y-1)^2$

$∴(x-3)^2+(2y-1)^2=0，则有x-3=0，2y-1=0$

$∴x=3，y=\frac 12$

$原式=\frac 23x×3\sqrt{x}+6x×\frac {\sqrt{xy}}x-y×\frac {\sqrt{xy}}y+x^2×\frac {\sqrt{x}}x=3x\sqrt{x}+5\sqrt{xy}$

$∴当x=3，y=\frac 12时，原式=9\sqrt{3}+5\sqrt{\frac 32}=9\sqrt{3}+\frac 52\sqrt{6}$

$解：∵x+y=-6，xy=4$

$∴x\lt 0，y\lt 0$

$∴\sqrt{\frac yx}+\sqrt{\frac xy}=\sqrt{\frac {-y}{-x}}+\sqrt{\frac {-x}{-y}}=\frac {\sqrt{xy}}{-x}+\frac {\sqrt{xy}}{-y}=-\frac {\sqrt{xy}(x+y)}{xy}$

$将x+y=-6，xy=4代入得$

$原式=-\frac {\sqrt{4}×(-6)}4=3$