电子课本网 第119页

第119页

信息发布者:
A
C
$解:原式​=3\sqrt {3}​$
$解:原式​=3\sqrt {2}-\sqrt {2}​$
$​ =2\sqrt {2}​$
$解:原式​=3\sqrt {ab}-2\sqrt {ab}+\frac {1}{3}\sqrt {ab}-\sqrt {ab}​$
$​ =\frac {1}{3}\sqrt {ab}​$
$解:原式​=10\sqrt {3}-4\sqrt {2}-9\sqrt {3}​$
$​ =\sqrt {3}-4\sqrt {2}​$
$解:原式​=6\sqrt {2}-5\sqrt {2}+\frac {1}{3}×3\sqrt {5}​$
$​ =\sqrt {2}+\sqrt {5}​$
$解:原式​=6\sqrt {5}+9\sqrt {5}-6\sqrt {5}​$
$​ =9\sqrt {5}​$
$解:原式​=\sqrt {3a}+3\sqrt {a}-\sqrt {3a}​$
$​ =3\sqrt {a}​$
$解:原式​=2\sqrt {3}-10\sqrt {2}-5\sqrt {3}+3\sqrt {2}​$
$​ =-3\sqrt {3}-7\sqrt {2}​$
$解:长:​\frac {1}{2}\sqrt {32}=\frac {1}{2}×4\sqrt {2}=2\sqrt {2}​$
$宽:​\frac {1}{3}\sqrt {18}=\frac {1}{3}×3\sqrt {2}=\sqrt {2}​$
$∴周长为​2(2\sqrt {2}+\sqrt {2})=6\sqrt {2}​$
$对角线长度​\sqrt {(2\sqrt {2})²+(\sqrt {2})²}=\sqrt {10}​$