Question

9．解方程组$\left\{\begin{array}{l}{{x}^{2}-2xy+{y}^{2}=1}\\{（x+y）^{2}-3（x+y）-10=0}\end{array}\right.$的最好方法是将两个方程都分解因式，那么原方程组可化为$\left\{\begin{array}{l}{x-y+1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y+1=0}\\{x+y-5=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y-5=0}\end{array}\right.$方程组来解．

Answer 1

分析 将两个方程分别因式分解可得（x-y+1）（x-y-1）=0且（x+y+2）（x+y-5）=0，继而可得．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-2xy+{y}^{2}=1}&{①}\\{（x+y）^{2}-3（x+y）-10=0}&{②}\end{array}\right.$，
由①可得（x-y）²-1=0，
（x-y+1）（x-y-1）=0，
∴x-y+1=0或x-y-1=0，
由②可得（x+y+2）（x+y-5）=0，
∴x+y+2=0或x+y-5=0，
∴原方程组可化为$\left\{\begin{array}{l}{x-y+1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y+1=0}\\{x+y-5=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y-5=0}\end{array}\right.$来解，
故答案为：$\left\{\begin{array}{l}{x-y+1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y+1=0}\\{x+y-5=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y+2=0}\end{array}\right.$或$\left\{\begin{array}{l}{x-y-1=0}\\{x+y-5=0}\end{array}\right.$．

点评 本题主要考查解高次方程的能力，解二元二次方程组的基本思想是“转化”，这种转化包含“消元”和“降次”将二元转化为一元是消元，将二次转化为一次是降次，这是转化的基本方法．因此，掌握好消元和降次的一些方法和技巧是解二元二次方程组的关键．