Question

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

Answer 1

分析 根据x²-8y²=0可得x=±2$\sqrt{2}$|y|，再分情况：①当x=2$\sqrt{2}$|y|，y＞0时；②当x=2$\sqrt{2}$|y|，y＜0时；③当x=-2$\sqrt{2}$|y|，y＞0时；④当x=-2$\sqrt{2}$|y|，y＜0时分别进行计算即可．

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

点评 此题主要考查了分式的值，在解答时应从已知条件和所求问题的特点出发，通过适当的变形、转化，才能发现解题的捷径．