Question

1.已知x^2+2x+1+y^2-4y+4=0,求x,y.

Answer 1

1.已知x^2+2x+1+y^2-4y+4=0,求x,y.

x^2+2x+1+y^2-4y+4=0

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

由于平方数都大于或等于0,所以必有

(x+1)^2=0,解得：x=-1,

(y-2)^2=0,解得：y=2；

2.求x^2+y^2-6x+4y+13的最小值.

x^2+y^2-6x+4y+13

=(x^2-6x+9)+(y^2+4y+4)

=(x-3)^2+(y+2)^2≥0

所以最小值为0.