Question

4．解方程組$\left\{\begin{array}{l}{{x}^{2}-2xy+{y}^{2}=9①}\\{（x+y）^{2}-5（x+y）+6=0②}\end{array}\right.$．

Answer 1

分析 分析：把方程①中用平方法轉化為兩個二元一次方程x-y=3或x-y=-3，方程②用因式分解法轉化為兩個二元一次方程x+y-2=0或x+y-3=0，最后原方程組轉化為四個二元一次方程組來求解．

解答 解：由方程①得x-y=±3，
∴x-y=3或x-y=-3
由方程②，得（x+y-2）（x+y-3）=0，
∴x+y-2=0或x+y-3=0，

∴原方程組可化為$\left\{\begin{array}{l}{x-y=3}\\{x+y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=3}\\{x+y-3=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-3}\\{x+y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-3}\\{x+y-3=0}\end{array}\right.$．

所以原方程組的解為：$\left\{\begin{array}{l}{x=\frac{5}{2}}\\{y=-\frac{1}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=3}\\{y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\frac{1}{2}}\\{y=\frac{5}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=3}\end{array}\right.$

點評 本題考查了高次方程．關鍵是將方程組中的某個方程左邊因式分解，使其積為0或兩邊開平方，可將較復雜的高次方程組轉化為簡單的高次方程組或者一次方程組來求解．