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第112页

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解:$\begin {aligned}\frac {\sqrt {2Rh_{1}}}{\sqrt {2Rh_{2}}}&=\frac {\sqrt {2Rh_{1}}}{\sqrt {2Rh_{2}}}\\&=\sqrt {\frac {h_{1}}{h_{2}}}\\&=\frac {\sqrt {h_{1}h_{2}}}{h_{2}}\end {aligned}$

解:(1)原式$=\frac {\sqrt {21}}{2\sqrt {6}}=\frac {\sqrt {14}}{4}$

(2)原式$=\sqrt {\frac {b^3}{8a^2}}=\frac {b\sqrt {2b}}{4a}$

(3)原式$=\sqrt {\frac {2(a - b)}{27(a + b)}}=\frac {\sqrt {6(a^2-b^2)}}{9(a + b)}$

(4)原式$=\frac {2(a + 2)}{2\sqrt {3(a + 2)}}=\frac {\sqrt {3(a + 2)}}{3}$

解$：(1)$原式$=\sqrt {9×2}$

$=\sqrt {9}×\sqrt {2}$

$=3\sqrt {2}$

解$：(2)$原式$=\frac {\sqrt {27}}{\sqrt {8}}$

$=\frac {3\sqrt {3}}{2\sqrt {2}}$

$=\frac {3\sqrt {6}}{4}$

解$：(3)$原式$=\frac {2\sqrt {}3}{2\sqrt {2x}}$

$=\frac {\sqrt {3}}{\sqrt {2x}}$

$=\frac {\sqrt {3}×\sqrt {2x}}{\sqrt {2x}×\sqrt {2x}}$

$=\frac {\sqrt {6x}}{2x}$

解$：(4)$原式$=\sqrt {\frac {1}{12}}$

$=\frac {\sqrt {12}}{12}$

$=\frac {2\sqrt {3}}{12}$

$=\frac {\sqrt {3}}{6}$

$ B$

解$：(1)$原式$=\frac {\sqrt {50}}{50}$

$=\frac {5\sqrt {2}}{50}$

$= \frac {\sqrt {2}}{10}$

解$：(2)$原式$=\sqrt {\frac {40}{9}}$

$=\frac {\sqrt {40}}{\sqrt {9}}$

$= \frac {2\sqrt {10}}{3}$

解$：(3)$原式$=\frac {\sqrt {3}}{\sqrt {8x}}$

$=\frac {\sqrt {3}}{2\sqrt {2x}}$

$= \frac {\sqrt {6x}}{4x}$

解$：(4)$原式$=\frac {\sqrt {0.04}}{\sqrt {0.28}×\sqrt {121}}$

$=\frac {1}{\sqrt {7}×11}$

$= \frac {\sqrt {7}}{77}$