Question

5．解方程组$\left\{\begin{array}{l}{{x}^{2}-{y}^{2}=5（x+y）}\\{{x}^{2}+xy+{y}^{2}=43}\end{array}\right.$．

Answer 1

分析 x²-y²=5（x+y），可得x+y=0或x-y=5．分别代入x²+xy+y²=43解出即可．

解答 解：∵x²-y²=5（x+y），∴x+y=0或x-y=5．
当x+y=0时，把y=-x代入x²+xy+y²=43可得x²=43，解得$\left\{\begin{array}{l}{x=\sqrt{43}}\\{y=-\sqrt{43}}\end{array}\right.$或$\left\{\begin{array}{l}{x=-\sqrt{43}}\\{y=\sqrt{43}}\end{array}\right.$．
当x-y=5时，把y=x-5代入x²+xy+y²=43可得x²-5x-6=，解得$\left\{\begin{array}{l}{x=6}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=-1}\\{y=-6}\end{array}\right.$．
综上可得原方程组的解为$\left\{\begin{array}{l}{x=\sqrt{43}}\\{y=-\sqrt{43}}\end{array}\right.$，$\left\{\begin{array}{l}{x=-\sqrt{43}}\\{y=\sqrt{43}}\end{array}\right.$，$\left\{\begin{array}{l}{x=6}\\{y=1}\end{array}\right.$，或$\left\{\begin{array}{l}{x=-1}\\{y=-6}\end{array}\right.$．

点评 本题考查了方程组的解法，考查了计算能力，属于中档题．