Question

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

Answer 1

分析 把a(bǔ)，b的值代入代數(shù)式，根據(jù)二次根式的性質(zhì)化簡(jiǎn)，即可解答．

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

點(diǎn)評(píng) 本題考查了二次根式的化簡(jiǎn)求值，解決本題的關(guān)鍵是熟記二次根式的乘除．