Question

16．解方程組：$\left\{\begin{array}{l}{{x}^{2}-xy+2x=0}\\{{x}^{2}-6xy+9{y}^{2}=4}\end{array}\right.$．

Answer 1

分析 由②得出x-3y=±2，由①得出x（x-y+2）=0，組成四個方程組，求出方程組的解即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}-xy+2x=0①}\\{{x}^{2}-6xy+9{y}^{2}=4②}\end{array}\right.$
由②得：（x-3y）²=4，
x-3y=±2，
由①得：x（x-y+2）=0，
x=0，x-y+2=0，
原方程組可以化為：$\left\{\begin{array}{l}{x=0}\\{x-3y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{x-3y=-2}\end{array}\right.$，$\left\{\begin{array}{l}{x-y+2=0}\\{x-3y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x-y+2=0}\\{x-3y=-2}\end{array}\right.$，
解得，原方程組的解為：$\left\{\begin{array}{l}{{x}_{1}=0}\\{{y}_{1}=-\frac{2}{3}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=0}\\{{y}_{2}=\frac{2}{3}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=-4}\\{{y}_{3}=-2}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-2}\\{{y}_{4}=0}\end{array}\right.$．

點評 本題考查了解高次方程組，能把高次方程組轉化二元一次方程組是解此題的關鍵．