Question

17．解方程組：$\left\{\begin{array}{l}{（{x}^{2}+1）（{y}^{2}+1）=10}\\{（x+y）（xy-1）=3}\end{array}\right.$．

Answer 1

分析 利用分解因式以及換元法，化簡方程組，然后求解方程組的解即可．

解答 解：方程組：$\left\{\begin{array}{l}{（{x}^{2}+1）（{y}^{2}+1）=10}\\{（x+y）（xy-1）=3}\end{array}\right.$，
化為：$\left\{\begin{array}{l}{（x+y）^{2}+（xy-1）^{2}=10}\\{（x+y）（xy-1）=3}\end{array}\right.$，
令a=x+y，b=xy-1．
可得$\left\{\begin{array}{l}{{a}^{2}+^{2}=10}\\{ab=3}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=1}\\{b=3}\end{array}\right.$，或$\left\{\begin{array}{l}{a=-1}\\{b=-3}\end{array}\right.$或$\left\{\begin{array}{l}{a=3}\\{b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=-3}\\{b=-1}\end{array}\right.$．
即：$\left\{\begin{array}{l}{x+y=1}\\{xy-1=3}\end{array}\right.$，解得方程組無解．
或$\left\{\begin{array}{l}{x+y=-1}\\{xy-1=-3}\end{array}\right.$，解得方程組無解．
或$\left\{\begin{array}{l}{x+y=3}\\{xy-1=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=2}\\{y=1}\end{array}\right.$或$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$
或$\left\{\begin{array}{l}{x+y=-3}\\{xy-1=-1}\end{array}\right.$，解得：$\left\{\begin{array}{l}{x=-3}\\{y=0}\end{array}\right.$或$\left\{\begin{array}{l}{x=0}\\{y=-3}\end{array}\right.$．

點評 本題考查曲線與方程的關系，化簡方程組，通過換元法求解的關鍵，考查計算能力．