第131页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

$\sqrt{21}$

$\sqrt{3n}$

$8\sqrt{2}$

$解:∵2\sqrt{3}=\sqrt{2²}×\sqrt{3}=\sqrt{4×3}= \sqrt{12}，$

$3\sqrt{2}=\sqrt{3²}×\sqrt{2}=\sqrt{9×2}=\sqrt{18}，$

$且12＜18，∴\sqrt{12}＜ \sqrt{18}，即2\sqrt{3}＜3\sqrt{2}.$

$解:∵4\sqrt{3}= \sqrt{4²}×\sqrt{3}= \sqrt{16×3}= \sqrt{48}，$

$3\sqrt{5}= \sqrt{3²}×\sqrt{5}= \sqrt{9×5}= \sqrt{45}，且48＞45，$

$∴ \sqrt{48}＞ \sqrt{45}，∴4\sqrt{3}＞3\sqrt{5}，∴-4\sqrt{3}＜-3\sqrt{5}$

$\sqrt{n+\frac {1}{n+2}}=(n+1)\sqrt{\frac{1}{n+2}}(n为正整数)$

$ \begin{aligned}解:原式&=2023 \sqrt{\frac{1}{2024}}× \sqrt{4048} \\ &=2023 \sqrt{\frac{1}{2024}×4048} \\ &=2023\sqrt{2}. \\ \end{aligned}$

（更多请点击查看作业精灵详解）

$ \begin{aligned}解:原式&=\frac {4}{3}\sqrt{2×72×\frac {5}{12}} \\ &=\frac {4}{3}\sqrt{60} \\ &=\frac {4}{3}\sqrt{2^2×15} \\ &=\frac {8\sqrt{15}}{3} \\ \end{aligned}$

$ \begin{aligned}解:∵x＜0，∴|x|&=-x \\ ∴原式&=\sqrt{x^2(x^2+1)} \\ &=|x|\sqrt{x^2+1} \\ &=-x\sqrt{x^2+1} \\ \end{aligned}$

$解:∵a＞0，b＞0$

$ \begin{aligned} ∴原式&=6\sqrt{8a}·(-2\sqrt{32ab}) \\ &=6\sqrt{8a×32ab} \\ &=6\sqrt{256a^2b} \\ &=6×16a\sqrt{b} \\ &=96a\sqrt{b} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {b}{b}\sqrt{3a}·\frac {1}{a}\sqrt{ab}·\frac {\sqrt{b}}{b} \\ &=\frac {\sqrt{3a}·\sqrt{ab}·\sqrt{b}}{ab} \\ &=\frac {\sqrt{3}ab}{ab} \\ &=\sqrt{3} \\ \end{aligned}$