Question

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

Answer 1

分析 先將原方程組變形為$\left\{\begin{array}{l}{（x-3y+3）（x-3y-3）=0}\\{（x+y-4）（x-y）=0}\end{array}\right.$，再變形為$\left\{\begin{array}{l}{x-3y+3=0}\\{x+y-4=0}\end{array}\right.，\left\{\begin{array}{l}{x-3y+3=0}\\{x-y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-3y-3=0}\\{x+y-4=0}\end{array}\right.，\left\{\begin{array}{l}{x-3y-3=0}\\{x-y=0}\end{array}\right.$，最后解這四個二元一次方程組求出其解即可．

解答 解：原方程組變形為：
$\left\{\begin{array}{l}{（x-3y+3）（x-3y-3）=0}\\{（x+y-4）（x-y）=0}\end{array}\right.$，
∴$\left\{\begin{array}{l}{x-3y+3=0}\\{x+y-4=0}\end{array}\right.，\left\{\begin{array}{l}{x-3y+3=0}\\{x-y=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-3y-3=0}\\{x+y-4=0}\end{array}\right.，\left\{\begin{array}{l}{x-3y-3=0}\\{x-y=0}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{{x}_{1}=\frac{9}{4}}\\{{y}_{1}=\frac{7}{4}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=\frac{3}{2}}\\{{y}_{2}=\frac{3}{2}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{3}=\frac{15}{4}}\\{{y}_{3}=\frac{1}{4}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-\frac{3}{2}}\\{{y}_{4}=-\frac{3}{2}}\end{array}\right.$．

點(diǎn)評 本題考查了二元一次方程組的解法的運(yùn)用，解二元二次方程組的消元、降次思想的運(yùn)用，因式分解的運(yùn)用．解答時(shí)先將二元高次方程變形為二元一次方程組是關(guān)鍵．