Question

15．已知x、y是整数，且x²-y²=7，则x^y=64或-$\frac{1}{64}$或-64或$\frac{1}{64}$．

Answer 1

分析 由平方差公式可知：（x+y）（x-y）=7，得出x+y=±1或±7，对应x-y=±7或±1，由此分别列出方程组解答即可．

解答 解：∵x²-y²=7，
∴（x+y）（x-y）=7，
∵x、y是整数，
∴x+y=±1或±7，对应x-y=±7或±1，
则$\left\{\begin{array}{l}{x+y=7}\\{x-y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x+y=-7}\\{x-y=-1}\end{array}\right.$或$\left\{\begin{array}{l}{x+y=-1}\\{x-y=-7}\end{array}\right.$或$\left\{\begin{array}{l}{x+y=1}\\{x-y=7}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=4}\\{y=3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-4}\\{y=-3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-4}\\{y=3}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=-3}\end{array}\right.$；
∴x^y=64或-$\frac{1}{64}$或-64或$\frac{1}{64}$．
故答案为：64或-$\frac{1}{64}$或-64或$\frac{1}{64}$．

点评 此题考查因式分解的实际运用，掌握平方差公式和分类讨论思想是解决问题的关键．