Question

1．已知实数x、y满足x²-$\sqrt{2}$xy-4y²=0，求$\frac{{x}^{2}-\sqrt{2}xy+{y}^{2}}{{x}^{2}+\sqrt{2}xy+{y}^{2}}$的值．

Answer 1

分析 首先将原式分解因式，进而得出x，y的关系，进而化简求出答案．

解答 解：∵x²-$\sqrt{2}$xy-4y²=0，
∴（x-2$\sqrt{2}$y）（x+$\sqrt{2}$y）=0，
∴x=2$\sqrt{2}$y，x=-$\sqrt{2}$y，
故当x=2$\sqrt{2}$y时，$\frac{{x}^{2}-\sqrt{2}xy+{y}^{2}}{{x}^{2}+\sqrt{2}xy+{y}^{2}}$=$\frac{4{y}^{2}+{y}^{2}}{（2\sqrt{2}y）^{2}+\sqrt{2}•2\sqrt{2}{y}^{2}+{y}^{2}}$=$\frac{5}{13}$，
当x=-$\sqrt{2}$y时，$\frac{{x}^{2}-\sqrt{2}xy+{y}^{2}}{{x}^{2}+\sqrt{2}xy+{y}^{2}}$=$\frac{5{y}^{2}}{（-\sqrt{2}y）^{2}+\sqrt{2}×（-\sqrt{2}y）•{y}^{2}+{y}^{2}}$=5．

点评 此题主要考查了二次根式的化简求值，正确得出x，y的关系是解题关键．