Lời giải:

`A=2x^2+y^2-2xy-4x-2y+2026`

`=x^2+y^2-2xy+2x-2y+x^2-6x+9+2016`

`=(x-y+1)^2+(x-3)^2+2016`

Có: \begin{cases}(x-y+1)^2\geq0\\(x-3)^2\geq0\end{cases}

`=> (x-y+1)^2+(x-3)^2>=0`

`=> (x-y+1)^2+(x-3)^2+2016>=2016`

Dấu "=" xảy ra `<=> x-y+1=0, x-3=0`

`=> x=3, y=4`

Vậy `A_(min)=2016 <=>` \begin{cases}x=3\geq0\\y=4\geq0\end{cases}

Đáp án:

 Min=2016

Giải thích các bước giải:

\(\begin{array}{l}
A = {x^2} + {\left( { - y} \right)^2} + 1 + 2x\left( { - y} \right) + 2.x.1 + 2.\left( { - y} \right).1 + {x^2} - 6x + 9 + 2016\\
 = {\left( {x - y + 1} \right)^2} + {\left( {x - 3} \right)^2} + 2016\\
Do:\left\{ \begin{array}{l}
{\left( {x - y + 1} \right)^2} \ge 0\\
{\left( {x - 3} \right)^2} \ge 0
\end{array} \right.\forall x;y\\
 \to {\left( {x - y + 1} \right)^2} + {\left( {x - 3} \right)^2} \ge 0\\
 \to {\left( {x - y + 1} \right)^2} + {\left( {x - 3} \right)^2} + 2016 \ge 2016\\
 \to Min = 2016\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - y + 1 = 0\\
x - 3 = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 3\\
y = 4
\end{array} \right.
\end{array}\)