已知 $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}},y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ ，求x²y²， $\sqrt{x}+\sqrt{y}$

答案

解： $x^{2}y^{2}=(xy)^{2}=1$
$x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=5-2\sqrt{6},y=5+2\sqrt{6}$
从而 $\sqrt{x}+\sqrt{y}$ = $\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{(\sqrt{3}-\sqrt{2})^{2}}+\sqrt{(\sqrt{3}+\sqrt{2})^{2}}$
= $\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}$