Question

3．已知a=$\frac{\sqrt{2}-1}{\sqrt{2}+1}$，b=$\frac{\sqrt{2}+1}{\sqrt{2}-1}$，求$\frac{b}{a}$+$\frac{a}{b}$的值．

Answer 1

分析 把a，b的值代入代数式，根据二次根式的性质化简，即可解答．

解答 解：$\frac{b}{a}+\frac{a}{b}$
=（$\frac{\sqrt{2}+1}{\sqrt{2}-1}$÷$\frac{\sqrt{2}-1}{\sqrt{2}+1}$）+（$\frac{\sqrt{2}-1}{\sqrt{2}+1}$÷$\frac{\sqrt{2}+1}{\sqrt{2}-1}$）
=$\frac{\sqrt{2}+1}{\sqrt{2}-1}×\frac{\sqrt{2}+1}{\sqrt{2}-1}$+$\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}+1}$
=$\frac{3+2\sqrt{2}}{3-2\sqrt{2}}$+$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
=$\frac{（3+2\sqrt{2}）^{2}-（3-2\sqrt{2}）^{2}}{（3-2\sqrt{2}）（3+2\sqrt{2}）}$
=$\frac{9+12\sqrt{2}+8-9+12\sqrt{2}-8}{9-8}$
=24$\sqrt{2}$．

点评 本题考查了二次根式的化简求值，解决本题的关键是熟记二次根式的乘除．