Lời giải:
$A=2x^2+3y^2-2xy+4x-7y+2012$
$2A=4x^2+6y^2-4xy+8x-14y+4024$
$=(4x^2-4xy+y^2)+5y^2+8x-14y+4024$
$=(2x-y)^2+4(2x-y)+5y^2-10y+4024$
$=(2x-y)^2+4(2x-y)+4+5(y^2-2y+1)+4015$
$=(2x-y+2)^2+5(y-1)^2+4015\geq 4015$
$\Rightarrow A\geq \frac{4015}{2}$
Vậy $A_{\min}=\frac{4015}{2}$ khi $(x,y)=(\frac{-1}{2},1)$