第130页 - 江苏版启东中学作业本八年级数学（上下册）

D

$2\sqrt{3}+10 \sqrt{2}$

0

$\frac{21}{4}$

$ \begin{aligned}解：原式&=2\sqrt{6}+\frac{\sqrt3}{3}-\frac{\sqrt3}{9}-6 \\ &=2\sqrt{6}+\frac{2}{9}\sqrt3-6 \\ \end{aligned}$

$ \begin{aligned}解：原式&=4\sqrt{6}+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{4}+\sqrt{6} \\ &=5\sqrt{6}+\frac{\sqrt{2}}{4} \\ \end{aligned}$

$ \begin{aligned}解：原式&=4 \sqrt{ab}+2\sqrt{ab}-(3\sqrt{ab}+3\sqrt{ab}) \\ &=6 \sqrt{ab}-6 \sqrt{ab} \\ &=0 \\ \end{aligned}$

$ \begin{aligned} 解：原式&=\sqrt{6}-\sqrt{2}+\sqrt{2}-1-3+\sqrt{6} \\ &=2\sqrt{6}-4 \\ \end{aligned}$

$ $

$\sqrt{n+1}-\sqrt{n}$

1

$解：原式=(\sqrt{2}-1+ \sqrt{3}- \sqrt{2}+···+ \sqrt{2024}- \sqrt{2023})( \sqrt{2024}+1)$

$=( \sqrt{2024}-1)( \sqrt{2024}+1)$

$=2024-1$

$=2023$