第113页 - 江苏版实验班提优训练数学八年级上下册答案

D

C

6

$ \begin{aligned} 解:原式&=\sqrt {4^{2}×9^{2} } \\ &=4×9 \\ &=36 \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\sqrt {2^{2} a^{2} b^{2} ·b} \\ &=2ab\sqrt {b} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\sqrt {\frac {2a}{5b}·\frac {2b}{c}·\frac {c}{5a}} \\ &=\sqrt {\frac {2×2}{5×5}} \\ &=\frac {2}{5} \\ \end{aligned}$

$ \begin{aligned}解:原式&= \sqrt{48}+ \sqrt{18} \\ &=4\sqrt{3}+3\sqrt{2} \\ \end{aligned}$

$ \begin{aligned}解:原式&=75+20 \sqrt{15}+20 \\ &=95+20 \sqrt{15} \\ \end{aligned}$

$ \begin{aligned}解:原式&=16-6 \sqrt{12} \\ &=16-12\sqrt{3} \\ \end{aligned}$

$ \begin{aligned} 解:原式&= (\sqrt{5})²-( \sqrt{3})² \\ &=5-3 \\ &=2 \\ \end{aligned}$

60

30x²

-20a²b

$解:∵x=\sqrt{2}-1，$

$∴x²+3x-1$

$=x²+2x+1+x-2 $

$= (x+1)²+x-2 $

$=( \sqrt{2}-1+1)²+\sqrt{2}-1-2 $

$=2+\sqrt{2}-3 $

$=-1+ \sqrt{2}. $

$解:∵a=3+\sqrt {2} ，b=3-\sqrt {2} ，\ $

$∴a+b=3+\sqrt{2}+3- \sqrt{2}=6，$

$a-b=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\ $

$则a²-b²=(a+b)(a-b)=6×2\sqrt{2}=12\sqrt{2}.$

$解:由(1)知a-b=2\sqrt{2}，\ $

$∴a²-2ab+b²=(a-b)²=(2\sqrt{2})²=8.$