电子课本网 第125页

第125页

信息发布者:
D
$\frac{1}{2}<x\leqslant 7$
$\frac{ab}{10} $
$ \begin{aligned} 解:原式&=\sqrt {\frac {n^2-\ \mathrm {m^2}}{\ \mathrm {m^2}n^2}} \\ &=\frac {\sqrt {n^2-\ \mathrm {m^2}}}{mn} \\ \end{aligned}$
$ \begin{aligned}解:原式&=\sqrt{\frac {3a^3×6b^3}{2ab}} \\ &=\sqrt{9a^2b^2} \\ &=3ab \\ \end{aligned}$
$ \begin{aligned}解:原式&=4÷(-5)×\sqrt{5÷\frac 95} \\ &=-\frac {4}5×\frac 53 \\ &=-\frac 43 \\ \end{aligned}$
$ \begin{aligned}解:原式&=\frac{3}{5}y^2\sqrt{xy}·(-\frac{15}{4}\sqrt{\frac{x}{y}})×(-\frac{5}{6}x\sqrt{xy}) \\ &=\frac{15}{8}x^2y^2 \sqrt{xy} \\ \end{aligned}$
$(1)证明:∵DF//AC,DE//BC$
$∴四边形ECFD是平行四边形$
$∵∠C=90°$
$∴四边形ECFD是矩形$
$(2)解:过点C作CH⊥EF 于点H,$
$则点C到EF 的距离即为CH的长$
$∵在Rt△ECF 中,CF=2,CE=4$
$∴EF= \sqrt {CE²+CF²} = \sqrt {20} =2 \sqrt {5}$
$∵S_{△ECF}= \frac {1}{2}\ \mathrm {CF} \cdot CE= \frac {1}{2}\ \mathrm {EF} \cdot CH$
$∴CH= \frac {CF \cdot CE}{EF} =\frac {2×4}{2\sqrt {5}} = \frac {4\sqrt {5}}{5}$
$∴点C到EF 的距离为 \frac {4\sqrt {5}}{5}$