Giải thích các bước giải:

Đặt `A=x^2+xy+y^2-3x-3y`

`=>2A=2x^2+2xy+2y^2-6x-6y`

`= (x^2+2xy+y^2)+(x^2-6x+9)+(y^2-6y+9)-18`

`=(x+y)^2+(x-3)^2+(y-3)^2-18>=-18`

`=> A>=-9`

Dấu '=' xảy ra `<=> x+y=0; x-3=0; y-3=0`

`=> x=y=3`

Vậy `A_(min) =-9` khi `x=y=3`

Đặt : `A=x^2+xy+y^2-3x-3y`

`⇒A=x^2+xy+y^2-2x-x-2y-y`

`⇒A=(x^2-2x+1)+(y^2-2y+1)+(xy-x-y+1)-3`

`⇒A=(x-1)^2+(y-1)^2+[x(y-1)-(y-1)]-3`

`⇒A=(x-1)^2+(y-1)^2+(x-1)(y-1)-3`

`⇒A=(x-1)^2+1/4(y-1)^2+3/4(y-1)^2+2 .1/2(x-1)(y-1)-3`

`⇒A=[(x-1)^2+2.1/2(x-1)(y-1)+1/4(y-1)^2]+3/4(y-1)^2-3`

`⇒A=[(x-1)+1/2(y-1)]^2+3/4(y-1)^2-3≥-3 ∀x`

Dấu $"="$ xảy ra khi :

$\begin{cases}[(x-1)+\dfrac12(y-1)]^2=0\\\dfrac34(y-1)^2=0\end{cases}$
$⇒\begin{cases}(x-1)+\dfrac12(y-1)=0\\y-1=0\end{cases}$

`⇒x=-1`