Question

4．方程组$\left\{\begin{array}{l}{x+xy+y=2+3\sqrt{2}}\\{{x}^{2}+{y}^{2}=6}\end{array}\right.$的解（x，y）=（2，$\sqrt{2}$）或（$\sqrt{2}$，2）．

Answer 1

分析 令x+y=a，xy=b，换元可求出a和b的值，再把a，b的值代入，联立方程组进行求解即可．

解答 解：x+y=a，xy=b，可得，
a+b=2+$3\sqrt{2}$，
a²-2b=6，
代入消去b，得：${a}^{2}+2a=10+6\sqrt{2}$，
解得：a=2+$\sqrt{2}$，b=2$\sqrt{2}$，或a=-4-$\sqrt{2}$，b=6+4$\sqrt{2}$，
当a=2+$\sqrt{2}$，b=2$\sqrt{2}$时，
$\left\{\begin{array}{l}{x+y=2+\sqrt{2}}\\{xy=2\sqrt{2}}\end{array}\right.$，
解得：$\left\{\begin{array}{l}{x=2}\\{y=\sqrt{2}}\end{array}\right.$，或$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=2}\end{array}\right.$，
当a=-4-$\sqrt{2}$，b=6+4$\sqrt{2}$时，
$\left\{\begin{array}{l}{x+y=-4-\sqrt{2}}\\{xy=6+4\sqrt{2}}\end{array}\right.$，
无解；
∴方程的解为：$\left\{\begin{array}{l}{x=2}\\{y=\sqrt{2}}\end{array}\right.$，$\left\{\begin{array}{l}{x=\sqrt{2}}\\{y=2}\end{array}\right.$，
故答案为：（2，$\sqrt{2}$）或（$\sqrt{2}$，2）

点评 此题主要考查高次方程的解法，会用换元思想进行降次是解题的关键．