Question

3．求方程2x²+xy-y²=14的整数解．

Answer 1

分析 首先把2x²+xy-y²=14左边分解为（2x-y）（x+y）=14，进一步把14分解成两个整数的乘积，进一步建立方程组得出答案即可．

解答 解：∵2x²+xy-y²=（2x-y）（x+y），14=1×14=（-1）×（-14）=2×7=（-2）×（-7），
∴$\left\{\begin{array}{l}{2x-y=1}\\{x+y=14}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=14}\\{x+y=1}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=-1}\\{x+y=-14}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=-14}\\{x+y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=2}\\{x+y=7}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=7}\\{x+y=2}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=-2}\\{x+y=-7}\end{array}\right.$，$\left\{\begin{array}{l}{2x-y=-7}\\{x+y=-2}\end{array}\right.$
解得$\left\{\begin{array}{l}{x=5}\\{y=9}\end{array}\right.$，$\left\{\begin{array}{l}{x=5}\\{y=-4}\end{array}\right.$，$\left\{\begin{array}{l}{x=-5}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=-5}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=4}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x=-3}\\{y=-4}\end{array}\right.$，$\left\{\begin{array}{l}{x=-3}\\{y=1}\end{array}\right.$．

点评 此题考查因式分解与二元一次方程组的实际运用，掌握因式分解的方法，注意分类讨论思想的运用．