第128页 - 江苏版启东中学作业本八年级数学（上下册）

（更多请查看作业精灵详解）

$ \begin{aligned}解：原式&=\frac{1}{3}×5 \sqrt{\frac{9}{4}×\frac{1}{28}×\frac{16}{7}} \\ &=\frac{5}{7} \\ \end{aligned}$

$ \begin{aligned} 解：原式&= -\frac{2}{y}×\frac{3}{2} × 3 \sqrt{xy^5·x^3y·\frac{x}{y}} \\ &= -\frac{9}{y}\sqrt{x^5y^5} \\ &=-9x^2y \sqrt{xy} \\ \end{aligned}$

$解：由秦九韶公式得：$

$ \begin{aligned} S&=\sqrt{\frac{1}{4}\left\{(\sqrt{7})^2×(2\sqrt{2})^2-[\frac {(\sqrt{7})^2+(2\sqrt{2})^2-3^3}2]^2 \right\}} \\ &=\sqrt{\frac{1}{4}[7×8-( \frac{7+8-9}{2} )^2]} \\ &= \sqrt{\frac{47}{4}} \\ &= \frac{\sqrt{47}}{2} \\ \end{aligned}$

$ $

（更多请查看作业精灵详解）

$ \begin{aligned}解：原式&=\frac{2}{3}× \sqrt{\frac{2}{45}}× \sqrt{\frac{5}{8}} \\ &=\frac{2}{3} \sqrt{\frac{2}{45}×\frac{5}{8}} \\ &=\frac{1}{9} \\ \end{aligned}$

$ \begin{aligned}原式&= \sqrt{3a^2}× \sqrt{\frac{2}{9a}}× \sqrt{\frac{a}{6}} \\ &= \sqrt{3a^2×\frac{2}{9a}×\frac{a}{6}} \\ &=\frac{a}{3} \\ \end{aligned}$

$ \begin{aligned} 原式&=3 \sqrt{3}÷(-\sqrt{6})×\frac{\sqrt6}{3} \\ &=3×(-1)×\frac{1}{3}×\sqrt{3×\frac{1}{6}×6} \\ &=- \sqrt{3} \\ \end{aligned}$

$ \begin{aligned}原式&=(3×\frac{1}{6}÷2) \sqrt{18×3÷6} \\ &=\frac{\sqrt9}{4} \\ &=\frac{3}{4} \\ \end{aligned}$

$解：根据题意，得1-4x≥0，4x-1≥0$

$∴x= \frac{1}{4}，则y= \frac{1}{2}$

$ 所以 \frac{x}{y}=\dfrac{\frac{1}{4}}{\frac{1}{2}}= \frac{1}{2} $

$ \frac{y}{x}=\dfrac{\frac{1}{2}}{\frac{1}{4}}=2 $

$所以 \sqrt{\frac{x}{y}+2+\frac{y}{x}} ÷ \sqrt{\frac{x}{y}-2+\frac{y}{x}}$

$ \begin{aligned} &=\sqrt{\frac{1}{2}+2+2} ÷ \sqrt{\frac{1}{2}-2+2} \\ &=\sqrt{\frac{9}{2}} ÷ \sqrt{\frac{1}{2}} \\ &=3 \\ \end{aligned}$

$∵p=\frac{7+8+9}{2}=12$

$∴由海伦公式得 $

$ \begin{aligned} S&=\sqrt{12×(12-7)×(12-8)×(12-9)} \\ &=\sqrt{12×5×4×3} \\ &=12\sqrt5 \\ \end{aligned}$