Question

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

Answer 1

分析 根据加减消元法，可得方程组的解．

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．
当时x=0时，y=±2，即方程组的解为$\left\{\begin{array}{l}{x=0}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$
当x=-y时，x=-y=$±\sqrt{2}$，$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，
综上所述：原方程组的解为$\left\{\begin{array}{l}{x=0}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=0}\\{y=-2}\end{array}\right.$，$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{2}}\\{y=-\sqrt{2}}\end{array}\right.$．

点评 本题考查了高次方程，加减消元法是解题的常用方法，因式分解将次是解题关键．