Question

5．求方程2x²-5xy+2y²=11的正整数解．

Answer 1

分析 根据因式分解将原方程变形为：（2x-y）（x-2y）=11，根据整数的整除性，可得：$\left\{\begin{array}{l}{2x-y=1}\\{x-2y=11}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=-1}\\{x-2y=-11}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=11}\\{x-2y=1}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=-11}\\{x-2y=-1}\end{array}\right.$；根据二元一次方程组的解法，求出x、y的值即可．

解答 解：2x²-5xy+2y²=11，
因式分解，得：（2x-y）（x-2y）=11，
即：$\left\{\begin{array}{l}{2x-y=1}\\{x-2y=11}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=-1}\\{x-2y=-11}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=11}\\{x-2y=1}\end{array}\right.$或$\left\{\begin{array}{l}{2x-y=-11}\\{x-2y=-1}\end{array}\right.$，
分别解二元一次方程组，得：$\left\{\begin{array}{l}{x=-3}\\{y=-7}\end{array}\right.$或$\left\{\begin{array}{l}{x=3}\\{y=7}\end{array}\right.$或$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$或$\left\{\begin{array}{l}{x=-7}\\{y=-3}\end{array}\right.$，
∵$\left\{\begin{array}{l}{x=-3}\\{y=-7}\end{array}\right.$或$\left\{\begin{array}{l}{x=-7}\\{y=-3}\end{array}\right.$不符合题意，
∴正整数解为：或$\left\{\begin{array}{l}{x=3}\\{y=7}\end{array}\right.$或$\left\{\begin{array}{l}{x=7}\\{y=3}\end{array}\right.$．

点评 本题主要考查因式分解、解二元一次方程组的综合应用，将方程因式分解，变形为二元一次方程组是解决此题的关键．