Question

1．解方程组：$\left\{\begin{array}{l}{{x}^{2}-4xy+4{y}^{2}=4，①}\\{x+y+1=0，②}\end{array}\right.$．

Answer 1

分析 由①得出x-2y=2或x-2y=-2，原方程组转化成两个二元一次方程组，求出方程组的解即可．

解答 解：由①得：x-2y=2或x-2y=-2．
原方程可化为$\left\{\begin{array}{l}x-2y=2\\ x+y=-1.\end{array}\right.$，$\left\{\begin{array}{l}x-2y=-2\\ x+y=-1.\end{array}\right.$
解得，原方程的解是$\left\{\begin{array}{l}{x_1}=0\\{y_1}=-1\end{array}\right.$，$\left\{\begin{array}{l}{x_2}=-\frac{4}{3}\\{y_2}=\frac{1}{3}.\end{array}\right.$．

点评 本题考查了解高次方程组，能把高次方程组转化成二元一次方程组是解此题的关键．