Question

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

Answer 1

分析 由②得出x²-3xy+2y²=0，即得x-2y=0，x-y=0，則原方程組可化為兩個(gè)二元一次方程組，求出方程組的解即可．

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，
則原方程組可化為
$\left\{\begin{array}{l}{x+2y=12}\\{x-2y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y=12}\\{x-y=0}\end{array}\right.$，
解這兩個(gè)方程組，得$\left\{\begin{array}{l}{{x}_{1}=6}\\{{y}_{1}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=4}\\{{y}_{2}=4}\end{array}\right.$．

點(diǎn)評 本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．