解方程(x2+y2)2+1=x2+y2+2xy

题目

解方程(x2+y2)2+1=x2+y2+2xy

答案

展开得 x^4+2x^2y^2+y^4+1=x^2+y^2+2xy ,
移项并分组得 (x^4-x^2+1/4)+(y^4-y^2+1/4)+2[(xy)^2-xy+1/4]=0 ,
分解因式得 (x^2-1/2)^2+(y^2-1/2)^2+2(xy-1/2)^2=0 ,
由于 (x^2-1/2)^2>=0 ,(y^2-1/2)^2>=0 ,(xy-1/2)^2>=0 ,
所以可得 x^2-1/2=0 ,y^2-1/2=0 ,xy-1/2=0 ,
解得 x=y=√2/2 或 x=y= -√2/2 .