`A=x^2-2xy+6y^2-12x+2y+45`
`A=(x^2-2xy+y^2)-12x+12y+5y^2-10y+45`
`A=(x-y)^2-12(x-y)+36+5(y^2-2y+1)+4`
`A=(x-y-6)^2+5(y-1)^2+4>=4`
Dấu = xảy ra khi $\begin{cases}x-y-6=0\\y-1=0\end{cases}$
$⇔\begin{cases}x=7\\y=1\end{cases}$
Vậy `A_(min)=4 <=> (x;y)=(7;1)`