Question

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

Answer 1

分析 先把方程组的两个方程变形，即可得出四个二元一次方程组，求出每个方程组的解即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+6xy+9{y}^{2}=9①}\\{{x}^{2}-{y}^{2}-4x+4y=0②}\end{array}\right.$
由①得：（x+y）²=9，
x+y=±3③，
由②得：（x+y）（x-y）-4（x-y）=0，
x-y=0，x+y-4=0，
即原方程组化为四个方程组：Ⅰ、$\left\{\begin{array}{l}{x+y=3}\\{x-y=0}\end{array}\right.$、Ⅱ、$\left\{\begin{array}{l}{x+y=3}\\{x+y-4=0}\end{array}\right.$，Ⅲ、$\left\{\begin{array}{l}{x+y=-3}\\{x-y=0}\end{array}\right.$，Ⅳ、$\left\{\begin{array}{l}{x+y=-3}\\{x+y-4=0}\end{array}\right.$
解得：方程组组Ⅰ的解为$\left\{\begin{array}{l}{x=1.5}\\{y=1.5}\end{array}\right.$，方程组Ⅱ无解，方程组Ⅲ的解为$\left\{\begin{array}{l}{x=-1.5}\\{y=-1.5}\end{array}\right.$，方程组Ⅳ无解，
所以原方程组的解为：$\left\{\begin{array}{l}{{x}_{1}=1.5}\\{{y}_{1}=1.5}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=-1.5}\\{{y}_{2}=-1.5}\end{array}\right.$．

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