Question

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

Answer 1

分析 由①得出x+2y=3，x+2y=-3③，由②得出x=0，x-y=0，則原方程組可化為四個二元一次方程組，求出方程組的解即可．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+4xy+4{y}^{2}=9①}\\{{x}^{2}-xy=0②}\end{array}\right.$
由①得：（x+2y）²=9，
即得x+2y=3，x+2y=-3③，
由②得：x（x-y）=0，
即x=0，x-y=0，
則原方程組可化為
$\left\{\begin{array}{l}{x+2y=3}\\{x=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y=3}\\{x-y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y=-3}\\{x=0}\end{array}\right.$，$\left\{\begin{array}{l}{x+2y=-3}\\{x-y=0}\end{array}\right.$，
解得，即可得出原方程組的解：$\left\{\begin{array}{l}{{x}_{1}=0}\\{{y}_{2}=6}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=1}\\{{y}_{2}=1}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=0}\\{{y}_{3}=-0.5}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-1}\\{{y}_{4}=-1}\end{array}\right.$．

點評 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．