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​$\sqrt {(2\sqrt {2})²+(\sqrt {2})²}=\sqrt {10}$