$\frac{\sqrt{2}+1}{\sqrt{3}+1}$ _________ $\frac{\sqrt{2}}{\sqrt{3}}$

答案

$\frac{\sqrt{2}+1}{\sqrt{3}+1}$ > $\frac{\sqrt{2}}{\sqrt{3}}$

$\frac{\frac{\sqrt{2}+1}{\sqrt{3}+1}}{\frac{\sqrt{2}}{\sqrt{3}}}=\frac{(\sqrt{2}+1)\sqrt{3}}{(\sqrt{3}+1)\sqrt{2}}=\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{2}}>1$ ,所以 $\frac{\sqrt{2}+1}{\sqrt{3}+1}$ > $\frac{\sqrt{2}}{\sqrt{3}}$

对二次根式的应用不熟悉

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