电子课本网 第111页

第111页

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解​$:(1)2\sqrt {12}=4\sqrt {3},\sqrt {27}=3\sqrt {3},$​是同类二次根式
​$(2)\sqrt {50}=5\sqrt {2},3\sqrt {8}=6\sqrt {2},$​是同类二次根式
​$(3)2\sqrt {1\frac {1}{2}}=\sqrt {6},2\sqrt {48}=8\sqrt {3},$​不是同类二次根式.
解​$:=(3\sqrt {5}+\sqrt {5})-(\sqrt {2}+4\sqrt {2})$​
​$ =4\sqrt {5}-5\sqrt {2}$​
解:原式​$=5\sqrt {3}-15\sqrt {3}-3\sqrt {3}$​
​$ =-13\sqrt {3}$​
解​$:=6\sqrt {2}+3\sqrt {2}-\frac {3\sqrt {2}}{2}$​
​$ =\frac {15\sqrt {2}}{2}$​
解:​$=4\sqrt {2}-4\sqrt {3}-\frac {\sqrt {3}}{3}-3\sqrt {2}$​
​$ =\sqrt {2}-\frac {13\sqrt {3}}{3}$​
解​$:=4\sqrt {3}-\frac {4\sqrt {3}}{9}+12\sqrt {3}$​
​$ =\frac {140\sqrt {3}}{9}$​
解:原式​$=\frac {\sqrt {2}}{2}-\frac {2\sqrt {3}}{3}-\frac {\sqrt {2}}{4}+5\sqrt {3}$​
​$ =\frac {\sqrt {2}}{4}+\frac {13\sqrt {3}}{3}$​