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第119页

信息发布者：

$-\frac {2\sqrt 5}{5}$

$-2$

$ \begin{aligned}解:原式&=\sqrt {\frac {49}{9}a^2×a} \\ &=\frac {7a}{3}\sqrt a \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt {\frac {9}{36}-\frac {4}{36}} \\ &=\sqrt {\frac {5}{36}} \\ &=\frac{\sqrt{5}}{6} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {(m-n)×\sqrt {m-n}}{\sqrt {m-n}×\sqrt {m-n}} \\ &=\sqrt {m-n} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\sqrt {x^4×\frac 2x} \\ &=\sqrt {x^2×2x} \\ &=x\sqrt{2x} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=−12\sqrt {48}÷12\sqrt 3 \\ &=-\sqrt {16} \\ &=−4 \\ \end{aligned}$

$ \begin{aligned} 解:原式&=-\sqrt {10}÷(-2\sqrt 5)×(-\frac {\sqrt 2}{2}) \\ &=\frac {\sqrt 2}{2}×(-\frac {\sqrt 2}{2}) \\ &=-\frac{1}{2} \\ \end{aligned}$

$解:原式=a^2+4ab+4b^2+a^2-4b^2+2ab-2a^2=6ab，$

$∵a=\frac{1}{\sqrt 3+\sqrt{2}}=\sqrt 3-\sqrt{2}，$

$b=\frac{1}{\sqrt 3-\sqrt{2}}=\sqrt 3+\sqrt{2}，$

$ \begin{aligned} ∴原式&=6×(\sqrt{3}-\sqrt{2})×(\sqrt{3}+\sqrt{2}) \\ &=6×[(\sqrt{3})^2-(\sqrt{2})^2] \\ &=6×(3-2) \\ &=6. \\ \end{aligned}$