Question

若x^2-4x+y^2+2y+5=0,求x,y的值

Answer 1

x^2-4x+y^2+2y+5=0

推出x^2-4x+y^2+2y+4+1=0

(x-2)^2+(y+1)^2=0

因为(x-2)^2>=0 +(y+1)^2>=0

所以只能(x-2)^2=0 (y+1)^2=0 即x-2=0 y+1=0

所以x=2且 y=-1