Question

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

Answer 1

分析 利用分解因式以及换元法，化简方程组，然后求解方程组的解即可．

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

点评 本题考查曲线与方程的关系，化简方程组，通过换元法求解的关键，考查计算能力．