Question

19．解方程组：$\left\{\begin{array}{l}{x+2y=12}\\{{x}^{2}+2{y}^{2}=3xy}\end{array}\right.$．

Answer 1

分析 由②得出x²-3xy+2y²=0，即得x-2y=0，x-y=0，则原方程组可化为两个二元一次方程组，求出方程组的解即可．

解答 解：$\left\{\begin{array}{l}{x+2y=12①}\\{{x}^{2}+2{y}^{2}=3xy②}\end{array}\right.$
由②，得x²-3xy+2y²=0，
即得x-2y=0，x-y=0，
则原方程组可化为
$\left\{\begin{array}{l}{x+2y=12}\\{x-2y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y=12}\\{x-y=0}\end{array}\right.$，
解这两个方程组，得$\left\{\begin{array}{l}{{x}_{1}=6}\\{{y}_{1}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=4}\end{array}\right.$．

点评 本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．