第119页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

A

C

$解：原式=3\sqrt {3}$

$解：原式=3\sqrt {2}-\sqrt {2}$

$ =2\sqrt {2}$

$解：原式=3\sqrt {ab}-2\sqrt {ab}+\frac {1}{3}\sqrt {ab}-\sqrt {ab}$

$ =\frac {1}{3}\sqrt {ab}$

$解：原式=10\sqrt {3}-4\sqrt {2}-9\sqrt {3}$

$ =\sqrt {3}-4\sqrt {2}$

$解：原式=6\sqrt {2}-5\sqrt {2}+\frac {1}{3}×3\sqrt {5}$

$ =\sqrt {2}+\sqrt {5}$

$解：原式=6\sqrt {5}+9\sqrt {5}-6\sqrt {5}$

$ =9\sqrt {5}$

$解：原式=\sqrt {3a}+3\sqrt {a}-\sqrt {3a}$

$ =3\sqrt {a}$

$解：原式=2\sqrt {3}-10\sqrt {2}-5\sqrt {3}+3\sqrt {2}$

$ =-3\sqrt {3}-7\sqrt {2}$

$解：长：\frac {1}{2}\sqrt {32}=\frac {1}{2}×4\sqrt {2}=2\sqrt {2}$

$宽：\frac {1}{3}\sqrt {18}=\frac {1}{3}×3\sqrt {2}=\sqrt {2}$

$∴周长为2(2\sqrt {2}+\sqrt {2})=6\sqrt {2}$

$对角线长度\sqrt {(2\sqrt {2})²+(\sqrt {2})²}=\sqrt {10}$