Question

9．已知：x=$\frac{1}{\sqrt{2}+1}$，y=$\frac{1}{\sqrt{2}-1}$，求值：x²+y²+xy．

Answer 1

分析 先对x、y进行化简，然后代入所求的式子即可解答本题．

解答 解：∵x=$\frac{1}{\sqrt{2}+1}$=$\sqrt{2}$-1，y=$\frac{1}{\sqrt{2}-1}$=$\sqrt{2}$+1，
∴x²+y²+xy
=$（\sqrt{2}-1）^{2}+（\sqrt{2}+1）^{2}+（\sqrt{2}-1）（\sqrt{2}+1）$
=2-$2\sqrt{2}$+1+2+2$\sqrt{2}$+1+2-1
=7．

点评 本题考查二次根式的化简求值，解题的关键是明确二次根式化简求值的方法．