Question

6．方程组$\left\{\begin{array}{l}{（x-y）（x+2y-1）=0}\\{（x+y-2）（x+y-1）=0}\end{array}\right.$ 可以转化为二元一次方程组$\left\{\begin{array}{l}{x-y=0}\\{x+y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x+y-1=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-2=0}\end{array}\right.$和$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-1=0}\end{array}\right.$．

Answer 1

分析 解原方程组中①得出x-y=0或x+2y-1=0；解原方程组中②得出x+y-2=0或x+y-1=0．方程①和②的解中各取一个结论联立成方程组即可得出结论．

解答 解：$\left\{\begin{array}{l}{（x-y）（x+2y-1）=0①}\\{（x+y-2）（x+y-1）=0②}\end{array}\right.$．
由①可得：x-y=0或x+2y-1=0；
由②可得：x+y-2=0或x+y-1=0．
∴方程组$\left\{\begin{array}{l}{（x-y）（x+2y-1）=0}\\{（x+y-2）（x+y-1）=0}\end{array}\right.$ 可以转化为二元一次方程组有$\left\{\begin{array}{l}{x-y=0}\\{x+y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x+y-1=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-2=0}\end{array}\right.$和$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-1=0}\end{array}\right.$．
故答案为：$\left\{\begin{array}{l}{x-y=0}\\{x+y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x+y-1=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-2=0}\end{array}\right.$和$\left\{\begin{array}{l}{x+2y-1=0}\\{x+y-1=0}\end{array}\right.$．

点评 本题考查了高次方程，将高次方程组转出成二元一次方程组是解题的关键．