x^2+y^2-2x+4y+7
=x^2-2x+1+y^2+4y+4+2
=(x-1)^2+(y+2)^2+2
(x-1)^2≥0
(y+2)^2≥0
(x-1)^2+(y+2)^2+2≥2
所以最小值是2