Question

17．若x²-8y²=0，則$\frac{x+y}{x-y}$=$\frac{9+4\sqrt{2}}{7}$或$\frac{9-4\sqrt{2}}{7}$．

Answer 1

分析 根據(jù)x²-8y²=0可得x=±2$\sqrt{2}$|y|，再分情況：①當(dāng)x=2$\sqrt{2}$|y|，y＞0時(shí)；②當(dāng)x=2$\sqrt{2}$|y|，y＜0時(shí)；③當(dāng)x=-2$\sqrt{2}$|y|，y＞0時(shí)；④當(dāng)x=-2$\sqrt{2}$|y|，y＜0時(shí)分別進(jìn)行計(jì)算即可．

解答 解：∵x²-8y²=0，
∴x²=8y²，
∴x=±2$\sqrt{2}$|y|，
①當(dāng)x=2$\sqrt{2}$|y|，y＞0時(shí)，$\frac{x+y}{x-y}$=$\frac{2\sqrt{2}y+y}{2\sqrt{2}y-y}$=$\frac{2\sqrt{2}+1}{2\sqrt{2}-1}$=$\frac{（2\sqrt{2}+1）^{2}}{（2\sqrt{2}-1）（2\sqrt{2}+1）}$=$\frac{9+4\sqrt{2}}{7}$；
②當(dāng)x=2$\sqrt{2}$|y|，y＜0時(shí)，$\frac{x+y}{x-y}$=$\frac{-2\sqrt{2}y+y}{-2\sqrt{2}y-y}$=$\frac{-2\sqrt{2}+1}{-2\sqrt{2}-1}$=$\frac{2\sqrt{2}-1}{2\sqrt{2}+1}$=$\frac{（2\sqrt{2}-1）^{2}}{（2\sqrt{2}+1）（2\sqrt{2}-1）}$=$\frac{9-4\sqrt{2}}{7}$；
③當(dāng)x=-2$\sqrt{2}$|y|，y＞0時(shí)，$\frac{x+y}{x-y}$=$\frac{-2\sqrt{2}y+y}{-2\sqrt{2}y-y}$=$\frac{-2\sqrt{2}+1}{-2\sqrt{2}-1}$=$\frac{2\sqrt{2}-1}{2\sqrt{2}+1}$=$\frac{（2\sqrt{2}-1）^{2}}{（2\sqrt{2}+1）（2\sqrt{2}-1）}$=$\frac{9-4\sqrt{2}}{7}$；
④當(dāng)x=-2$\sqrt{2}$|y|，y＜0時(shí)，$\frac{x+y}{x-y}$=$\frac{2\sqrt{2}y+y}{2\sqrt{2}y-y}$=$\frac{2\sqrt{2}+1}{2\sqrt{2}-1}$=$\frac{（2\sqrt{2}+1）^{2}}{（2\sqrt{2}-1）（2\sqrt{2}+1）}$=$\frac{9+4\sqrt{2}}{7}$；
故答案為：$\frac{9+4\sqrt{2}}{7}$或$\frac{9-4\sqrt{2}}{7}$．

點(diǎn)評(píng) 此題主要考查了分式的值，在解答時(shí)應(yīng)從已知條件和所求問(wèn)題的特點(diǎn)出發(fā)，通過(guò)適當(dāng)?shù)淖冃�、轉(zhuǎn)化，才能發(fā)現(xiàn)解題的捷徑．