电子课本网 第146页

第146页

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$3-\sqrt {3}\ $
$ \begin{aligned}解:原式&=3\sqrt {2} -\frac {3\sqrt {2} }{2}+2\sqrt {2} \\ &=\frac{7\sqrt {2} }{2}. \\ \end{aligned}$
$ \begin{aligned}解:原式&= \sqrt{3a²} × \sqrt{\frac{2}{9a}}×\sqrt{\frac{a}{6}} \\ &= \sqrt{3a²×\frac{2}{9a}×\frac{a}{6}} \\ &=\frac{a}{3}. \\ \end{aligned}$
$ \begin{aligned}解:原式&=5-3-(12-4\sqrt {6} +2) \\ &=2-14+4\sqrt {6} \\ &=4\sqrt {6} -12. \\ \end{aligned}$
$ \begin{aligned} 解:原式&=-1+4-\frac {\sqrt {3} }{3}×4\sqrt {3} +\sqrt {2} -1+1 \\ &=-1+4-4+\sqrt {2} \\ &=\sqrt{2}-1. \\ \end{aligned}$
$ \begin{aligned}解:原式&=\frac{2}{3}×3\sqrt{x}+6×\frac{\sqrt{x}}{2}-2x·\frac{\sqrt{x}}{x} \\ &=2\sqrt{x}+3\sqrt{x}-2\sqrt{x} \\ &=3\sqrt {x} , \\ \end{aligned}$
$取x>0 的值即可,$
$如:当x=4时,原式=3\sqrt{4}=6.$
$解:(1)设长方体的长为4x,宽为2x,高为\sqrt {5} x,$
$由题意得4x·2x=20, x=\frac{\sqrt {10} }{2},$
$∴高为\sqrt{5}×\frac{\sqrt{10}}{2}=\frac{5\sqrt {2} }{2}.$
$(2)此长方体的长为4× \frac{\sqrt{10}}{2} =2 \sqrt{10} ,$
$宽为2× \frac{\sqrt {10} }{2} = \sqrt{10} ,$
$表面积为2(20+2 \sqrt{10}× \frac {5\sqrt {2} }{2}+\sqrt{10} × \frac{5\sqrt {2} }{2} )$
$=40+30\sqrt {5} .$