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第134页

信息发布者：

$\frac{4}{3}\sqrt{3}$

$20\sqrt{3}$

$-\sqrt{3} + 4\sqrt{2}$

解：原式$=2\sqrt {2}+2\sqrt {3}-3\sqrt {3}-\sqrt {2}$

$ =\sqrt {2}-\sqrt {3}$

解：原式$ =3\sqrt {6}-\frac {\sqrt {6}}{3}-\sqrt {2}+4\sqrt {2}$

$ =\frac {8\sqrt {6}}{3}+3\sqrt {2}$

解：原式$ =-3×\frac {2\sqrt {6}}{9}×\frac {2}{\sqrt {3}}×3\sqrt {3}$

$ =-4\sqrt {6}$

解：原式$ =12-4\sqrt {3}+1+3-4$

$ =12-4\sqrt {3}$

解：原式$ =(-2\sqrt {6})^2-(3\sqrt {2})^2$

$ =24-18$

$ =6$

解：原式$ =[(2\sqrt {7}+5\sqrt {2})+(2\sqrt {7}-5\sqrt {2})][(2\sqrt {7}+5\sqrt {2})-(2\sqrt {7}-5\sqrt {2})]$

$ =4\sqrt {7}×10\sqrt {2}$

$ =40\sqrt {14}$

 解：

 $\because x=\sqrt{3}+\sqrt{7}, y=\sqrt{3}-\sqrt{7}$

 $\therefore x-y=2\sqrt{7}, xy=(\sqrt{3})^2-(\sqrt{7})^2=3-7=-4$

 $\therefore 4x^2-7xy+4y^2$

 $=4(x^2+y^2)-7xy$

 $=4[(x-y)^2+2xy]-7xy$

 $=4(x-y)^2+8xy-7xy$

 $=4(x-y)^2+xy$

 $=4×(2\sqrt{7})^2+(-4)$

 $=4×28-4$

 $=112-4$

 $=108$