电子课本网 第8页

第8页

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解​$:(1)$​原式​$=\frac {1}{3}+0.3-\frac {1}{3} $​
​$=0.3 ; $​
​$(2)$​原式​$=\sqrt {6}-\frac {1}{2} \sqrt {2}-2 \sqrt {6}-\frac {2}{3} \sqrt {6} $​
​$=-\frac {5}{3} \sqrt {6}-\frac {1}{2} \sqrt {2} ;$​
​$(3)$​原式​$= {[\sqrt {3}+\sqrt {5}+\sqrt {6}-(\sqrt {3}-\sqrt {5}+\sqrt {6})]·} $​
​$ {[\sqrt {3}+\sqrt {5}+\sqrt {6}+(\sqrt {3}-\sqrt {5}+\sqrt {6})] } $​
​$= 4 \sqrt {5}(\sqrt {3}+\sqrt {6}) $​
​$= 4 \sqrt {15}+4 \sqrt {30}$​

解:∵​$x=\frac {1}{2}(3+\sqrt {3}), y=\frac {1}{2}(3-\sqrt {3}), $​
∴​$x+y=3, x y=\frac {1}{4}(9-3)=\frac {3}{2}, $​
∴​$x^2+x y+y^2=(x+y)^2-x y=9 -\frac {3}{2}=\frac {15}{2}$​