第137页 - 亮点给力提优课时作业本八年级数学江苏版

$10\sqrt {2}$

$=9×\frac {3}{20}× \sqrt {45×\frac {5}{9}×\frac {12}{5}}$

$=\frac {27}{10}\sqrt {15}$

解：∵$a≥b≥0$，∴$a≥0$，$a-b≥0$

∴原$ $式$= \sqrt {16ab · a(a-b)²}=4a(a-b)\sqrt {b}$

解：设该直角三角形的面积为$S$

分情况讨论如下：$ $当$3\sqrt {2}$和$2\sqrt {3}$为该直角三角形的两条直角边长时，

$S=\frac {1}{2}×3\sqrt {2}×2\sqrt {3}=3\sqrt {6}$；

当$3\sqrt {2}$为该直角三角形的斜边长时，由勾股定理，得另一直角边长为$ \sqrt {(3\sqrt {2})²-(2\sqrt {3})²}= \sqrt {6}$

∴$S=\frac {1}{2}×2\sqrt {3}×\sqrt {6}=3\sqrt {2}$

综上，该直角三角形的面积为$3\sqrt {6}$或$3\sqrt {2}$

D

4

解：$(1)$∵$(\sqrt {30-x}+ \sqrt {9-x})(\sqrt {30-x}- \sqrt {9-x})=(\sqrt {30-x})²-(\sqrt {9-x})²=30-x-9+x=21$

且$ \sqrt {30-x}+ \sqrt {9-x}=7$

∴$\sqrt {30-x}- \sqrt {9-x}=3$

$(2)$由$(1)$及题意得$ \sqrt {30-x}- \sqrt {9-x}=3$，$\sqrt {30-x}+ \sqrt {9-x}= 7$

∴两式相加得$2 \sqrt {30-x}=10$，即$ \sqrt {30-x}=5$

∴$30-x=25$，解得$x=5$

则$x$的值为$5$