Question

20．解方程組：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}+x+y=18}\\{{x}^{2}+xy+{y}^{2}=19}\end{array}\right.$．

Answer 1

分析 兩式相減得到xy-x-y=1即x+y=xy-1，代入②式變形得到（xy-1）²-xy=19，解得xy=6或-3 當xy=6時，得到$\left\{\begin{array}{l}{xy=6}\\{x+y=xy-1}\end{array}\right.$，當xy=-3時，得到$\left\{\begin{array}{l}{xy=-3}\\{x+y=xy-1}\end{array}\right.$，再分別解方程組即可求解．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+{y}^{2}+x+y=18①}\\{{x}^{2}+xy+{y}^{2}=19②}\end{array}\right.$
②-①得：xy-x-y=1即x+y=xy-1③，
由②得：（x+y）²-xy=19④
把x+y=xy-1代入④得：（xy-1）²-xy=19，
整理得：（xy）²-3xy-18=0，
解得：xy=6或-3，
當xy=6時，則$\left\{\begin{array}{l}{xy=6}\\{x+y=xy-1}\end{array}\right.$
解得$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=3}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=3}\\{{y}_{2}=2}\end{array}\right.$
當xy=-3時，則$\left\{\begin{array}{l}{xy=-3}\\{x+y=xy-1}\end{array}\right.$
解得$\left\{\begin{array}{l}{{x}_{3}=-2-\sqrt{7}}\\{{y}_{3}=-2+\sqrt{7}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=-2+\sqrt{7}}\\{{y}_{4}=-2-\sqrt{7}}\end{array}\right.$．

點評 本題考查了解高次方程組，注意：解方程組的方法有兩種：①加減消元法，②代入消元法．