Question

7．求方程2xy-x-y-3=0的整数解．

Answer 1

分析 先把方程变形为：（2x-1）（2y-1）=7，根据x，y为整数和（2x-1）（2y-1）=7=1×7=7×1=-1×（-7）=-7×（-1），得到方程组$\left\{\begin{array}{l}{2x-1=1}\\{2y-1=7}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=7}\\{2y-1′=1}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=-1}\\{2y-1=-7}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=-7}\\{2y-1=-1}\end{array}\right.$，得到原方程的四组解．

解答 解：∵2xy-x-y-3=0，
∴4xy-2x-2y-6=0
∴4xy-2x-2y+1=7，
∴（2x-1）（2y-1）=7
∵x，y为整数，
∴（2x-1）与（2y-1）也是整数，
而（2x-1）（2y-1）=7=1×7=7×1=-1×（-7）=-7×（-1），
∴$\left\{\begin{array}{l}{2x-1=1}\\{2y-1=7}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=7}\\{2y-1′=1}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=-1}\\{2y-1=-7}\end{array}\right.$或$\left\{\begin{array}{l}{2x-1=-7}\\{2y-1=-1}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x=1}\\{y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x=4}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=0}\\{y=-3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-3}\\{y=0}\end{array}\right.$．

点评 本题是非一次不定方程（组），主要考查了求二元二次方程整数解的方法．代数式的变形能力以及整数的性质．等式的性质，分解因式，解本题的关键是把原方程变形为4xy-2x-2y+1=7之后的分解因式，难点是方程两边同时乘以2．