Question

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

Answer 1

分析 由①得出x-2y=2或x-2y=-2，原方程組轉(zhuǎn)化成兩個(gè)二元一次方程組，求出方程組的解即可．

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

點(diǎn)評(píng) 本題考查了解高次方程組，能把高次方程組轉(zhuǎn)化成二元一次方程組是解此題的關(guān)鍵．