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(zhuǎn)化為兩個(gè)二元一次方程x-y=3或x-y=-3，方程②用因式分解法轉(zhuǎn)化為兩個(gè)二元一次方程x+y-2=0或x+y-3=0，最后原方程組轉(zhuǎn)化為四個(gè)二元一次方程組來(lái)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(diǎn)評(píng) 本題考查了高次方程．關(guān)鍵是將方程組中的某個(gè)方程左邊因式分解，使其積為0或兩邊開(kāi)平方，可將較復(fù)雜的高次方程組轉(zhuǎn)化為簡(jiǎn)單的高次方程組或者一次方程組來(lái)求解．