第41页 - 启东中学作业本八年级数学苏科版宿迁专版

$\sqrt{ab}$

≥

$\sqrt{ab}$

≥

能开得尽方

$ \begin{aligned}解:原式&=\sqrt {16} \\ &=4 \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt {144} \\ &=12 \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt {16} \\ &=4 \\ \end{aligned}$

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

$解:原式=\sqrt {36}-5$

$\hspace{1.4cm}=6-5$

$\hspace{1.4cm}=1$

$ \begin{aligned}解:原式&=\sqrt {36}+\sqrt {100} \\ &=6+10 \\ &=16 \\ \end{aligned}$

$解:原式=14$

$ \begin{aligned}解:原式&=\sqrt {25}×\sqrt 2 \\ &=5\sqrt{2} \\ \end{aligned}$

$ 解:原式=\sqrt {6^2}×\sqrt {7^2} $

$\hspace{1.4cm}=6×7 $

$\hspace{1.4cm}=42 $

$解:原式=\sqrt {9x^2}•\sqrt y^4•\sqrt y$

$\hspace{1.4cm}=3|x|y^2\sqrt y$

$解:原式=2\sqrt{xy}$

$ \begin{aligned}解:原式&=\sqrt {16}•\sqrt {x^2} \\ &=4x \\ \end{aligned}$

$解:∵\sqrt{x+1}•\sqrt{2-x}= \sqrt{(x+1)(2-x)}成立，$

$∴\begin{cases}{\ x+1≥0，\ }\\{2-x≥0，}\end{cases}$

$解得-1≤x≤ 2.\ $