Question

1．解方程组：$\left\{\begin{array}{l}{{x}^{2}+3xy+{y}^{2}+2x+2y=8}\\{2{x}^{2}+2{y}^{2}+3x+3y=14}\end{array}\right.$．

Answer 1

分析 两式相减得到6xy+x+y=2即x+y=2-6xy，代入式变形得到（x+y）²+xy+2（x+y）=8，解得xy=0或$\frac{35}{36}$，当xy=0时，得到$\left\{\begin{array}{l}{xy=0}\\{x+y=2-6xy}\end{array}\right.$，当xy=$\frac{35}{36}$时，得$\left\{\begin{array}{l}{xy=\frac{35}{36}}\\{x+y=2-6xy}\end{array}\right.$，再分别解方程组即可求解．

解答 解：$\left\{\begin{array}{l}{{x}^{2}+3xy+{y}^{2}+2x+2y=8①}\\{2{x}^{2}+2{y}^{2}+3x+3y=14②}\end{array}\right.$
①×2-②得：6xy+x+y=2即x+y=2-6xy③，
由①得：（x+y）²+xy+2（x+y）=8④
把x+y=2-6xy代入④得：（2-6xy）²+xy+2（2-6xy）=8，
整理得：36x²y²-35xy=0，
解得：xy=0或$\frac{35}{36}$，
当xy=0时，则$\left\{\begin{array}{l}{xy=0}\\{x+y=2-6xy}\end{array}\right.$
解得$\left\{\begin{array}{l}{{x}_{1}=2}\\{{y}_{1}=0}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=0}\\{{y}_{2}=2}\end{array}\right.$，
当xy=$\frac{35}{36}$时，则$\left\{\begin{array}{l}{xy=\frac{35}{36}}\\{x+y=2-6xy}\end{array}\right.$，
解得$\left\{\begin{array}{l}{{x}_{3}=\frac{23+\sqrt{389}}{12}}\\{{y}_{3}=\frac{23-\sqrt{389}}{12}}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{4}=\frac{23-\sqrt{389}}{12}}\\{{y}_{4}=\frac{23+\sqrt{389}}{12}}\end{array}\right.$．

点评 本题考查了高次方程，二元方程组无论多复杂，解二元方程组的基本思想都是消元，消元的方法有代入法和加减法．