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第31页

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$解：原式=(6×\frac {\sqrt{x}}2-2x×\frac {\sqrt{x}}x)÷(-\frac 13\sqrt{x})$

$=(3\sqrt{x}-2\sqrt{x})÷(-\frac 13\sqrt{x})$

$=\sqrt{x}÷(-\frac 13\sqrt{x})$

$=-3$

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

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

$=\frac {4-6}{9}$

$=-\frac 29$

$ \begin{aligned}解：原式&=4x^2+4xy+y^2+x^2-y^2-5x^2+5xy \\ &=9xy \\ \end{aligned}$

$当x=\sqrt {6}-1，y=\sqrt {6}+1时,$

$原式=9×(\sqrt {6}-1)×(\sqrt {6}+1)=9×(6-1)=45$

$ \begin{aligned}解：原式&=\frac{a-1}{a+1}·\frac{a+1}{(a-1)^2} \\ &=\frac{1}{a-1}. \\ \end{aligned}$

$当a= \sqrt{2}+1时，原式=\frac{1}{\sqrt{2}+1-1}=\frac{\sqrt{2}}{2}$

$解：由题意得\begin{cases}3x-1≥0\\1-3x≥0\end{cases}$

$解得x=\frac 13$

$∴y=\frac 12$

$∴原式=\sqrt{xy}÷\frac {\sqrt{y}}{2y}=\sqrt{xy} \cdot \frac {2y}{\sqrt{y}}=2y\sqrt{x}=\frac {\sqrt3}3$