Question

20．解方程組：$\left\{\begin{array}{l}{3x^2-y^2=8}\\{x^2+xy+y^2=4}\end{array}\right.$．

Answer 1

分析 根據加減消元法，可得x²-2xy-3y²=0，根據因式分解，可得x、y的關系．

解答 解：$\left\{\begin{array}{l}{3{x}^{2}-{y}^{2}=8①}\\{{x}^{2}+xy+{y}^{2}=4②}\end{array}\right.$
①-②×2，得
x²-2xy-3y²=0．
因式分解，得
（x+y）（x-3y）=0．
解得x=-y，或x=3y．
當x=-y時，x=-y=±2，$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$，
當x=3y時，y=$±\frac{2\sqrt{13}}{13}$，$\left\{\begin{array}{l}{x=\frac{6\sqrt{13}}{13}}\\{y=\frac{2\sqrt{13}}{13}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{6\sqrt{13}}{13}}\\{y=-\frac{2\sqrt{13}}{13}}\end{array}\right.$，
原方程組的解為$\left\{\begin{array}{l}{x=2}\\{y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x=-2}\\{y=2}\end{array}\right.$，$\left\{\begin{array}{l}{x=\frac{6\sqrt{13}}{13}}\\{y=\frac{2\sqrt{13}}{13}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\frac{6\sqrt{13}}{13}}\\{y=-\frac{2\sqrt{13}}{13}}\end{array}\right.$．

點評 本題考查了高次方程，加減消元是解題常用方法，因式分解是解題關鍵．