Question

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

Answer 1

分析 根據加減消元法，可得方程組的解．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}=4}\\{xy-{y}^{2}+4=0}\end{array}\right.$，
①+②得x²+xy=0
因式分解，得x（x+y）=0
解得x=0或y=-x．
當時x=0時，y=±2，即方程組的解為$\left\{\begin{array}{l}{x=0}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$
當x=-y時，x=-y=$±\sqrt{2}$，$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，
綜上所述：原方程組的解為$\left\{\begin{array}{l}{x=0}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$，$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$．

點評 本題考查了高次方程，加減消元法是解題的常用方法，因式分解將次是解題關鍵．