第46页

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法则
公式
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乘方
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B
A
60
$10+2\sqrt{21}$
$10\sqrt{3}$
解：​$\sqrt {12}-\sqrt {6}×\frac {1}{\sqrt {2}}$​
​$=2\sqrt {3}-\sqrt {\frac {6}{2}}$​
​$=2\sqrt {3}-\sqrt {3}$​
​$=\sqrt {3}$​
解：​$(5+2\sqrt {6})(2\sqrt {6}-5)$​
​$=(2\sqrt {6})^2-5^2$​
​$=24-25$​
​$=-1$​
解：​$(\sqrt {20}+\sqrt {5})÷\sqrt {5}-\sqrt {\frac {1}{3}}×\sqrt {12}$​
​$=\sqrt {20}÷\sqrt {5}+\sqrt {5}÷\sqrt {5}-\sqrt {\frac {1}{3}×12}$​
​$=2+1-\sqrt {4}$​
​$=3-2$​
​$=1$​
解：​$(\sqrt {10}-\sqrt {3})(\sqrt {3}+\sqrt {10})-(\sqrt {2}-\sqrt {6})^2$​
​$=(\sqrt {10})^2-(\sqrt {3})^2-[(\sqrt {2})^2-2\sqrt {2}×\sqrt {6}+(\sqrt {6})^2]$​
​$=10-3-(2-4\sqrt {3}+6)$​
​$=7-(8-4\sqrt {3})$​
​$=7-8+4\sqrt {3}$​
​$=-1+4\sqrt {3}$​
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