第43页 - 启东中学作业本八年级数学苏科版宿迁专版

$\sqrt{\dfrac{a}{b}}$

≥

＞

$\sqrt{\dfrac{a}{b}}$

≥

＞

D

$ \begin{aligned}解:原式&=\sqrt 4 \\ &=2 \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt 8 \\ &=2\sqrt{2} \\ \end{aligned}$

$ 解:原式=2\sqrt 6×\frac {3}{2\sqrt 3} $

$\hspace{1.4cm}=\frac {3\sqrt 6}{\sqrt 3} $

$\hspace{1.4cm}=3\sqrt{2} $

$ \begin{aligned}解:原式&=\sqrt {128} \\ &=\sqrt {64×2} \\ &=8\sqrt{2} \\ \end{aligned}$

$解:原式=\frac{11}{12}$

$解:原式=\frac{\sqrt{5}}{3}$

$ \begin{aligned}解:原式&=\frac {\sqrt {18}}{7} \\ &=\frac{3\sqrt{2}}{7} \\ \end{aligned}$

$ 解:原式=\sqrt {\frac {124}{25}} $

$\hspace{1.4cm}=\frac {\sqrt {4×31}}{5} $

$\hspace{1.4cm}=\frac{2\sqrt{31}}{5} $

$ \begin{aligned}解:原式&=\sqrt {9} \\ &=3 \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt 8 \\ &=2\sqrt{2} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {6×5\sqrt 3}{2\sqrt 3} \\ &=\frac {90}{6} \\ &=15 \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {ab\sqrt a}{b} \\ &=a\sqrt{a} \\ \end{aligned}$