Question

10．已知x=$\frac{1}{\sqrt{2}-1}$，y=$\frac{1}{\sqrt{2}+1}$，求x²+3xy+y²的值．

Answer 1

分析 首先将x，y的值化简，进而代入原式结合完全平方公式求出即可．

解答 解：∵x=$\frac{1}{\sqrt{2}-1}$，y=$\frac{1}{\sqrt{2}+1}$，
∴x=$\frac{\sqrt{2}+1}{（\sqrt{2}-1）（\sqrt{2}+1）}$=$\sqrt{2}$+1，y=$\frac{\sqrt{2}-1}{（\sqrt{2}+1）（\sqrt{2}-1）}$=$\sqrt{2}$-1，
∴x²+3xy+y²
=（$\sqrt{2}$+1）²+3（$\sqrt{2}$+1）（$\sqrt{2}$-1）+（$\sqrt{2}$-1）²
=3+2$\sqrt{2}$+3+3-2$\sqrt{2}$
=9．

点评 此题主要考查了二次根式的化简求值，正确分母有理化是解题关键．