Question

1．已知實(shí)數(shù)x、y滿足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 首先將原式分解因式，進(jìn)而得出x，y的關(guān)系，進(jìn)而化簡(jiǎn)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，
故當(dāng)x=2$\sqrt{2}$y時(shí)，$\frac{{x}^{2}-\sqrt{2}xy+{y}^{2}}{{x}^{2}+\sqrt{2}xy+{y}^{2}}$=$\frac{4{y}^{2}+{y}^{2}}{（2\sqrt{2}y）^{2}+\sqrt{2}•2\sqrt{2}{y}^{2}+{y}^{2}}$=$\frac{5}{13}$，
當(dāng)x=-$\sqrt{2}$y時(shí)，$\frac{{x}^{2}-\sqrt{2}xy+{y}^{2}}{{x}^{2}+\sqrt{2}xy+{y}^{2}}$=$\frac{5{y}^{2}}{（-\sqrt{2}y）^{2}+\sqrt{2}×（-\sqrt{2}y）•{y}^{2}+{y}^{2}}$=5．

點(diǎn)評(píng) 此題主要考查了二次根式的化簡(jiǎn)求值，正確得出x，y的關(guān)系是解題關(guān)鍵．