Lời giải:
Ta có:
\(A=2x^2+y^2-2x-2xy+2017\)
\(=(x^2+y^2-2xy)+(x^2-2x+1)+2016\)
\(=(x-y)^2+(x-1)^2+2016\)
Ta thấy \((x-y)^2\geq 0; (x-1)^2\geq 0, \forall x,y\)
\(\Rightarrow A\geq 0+0+2016=2016\)
Vậy \(A_{\min}=2016\Leftrightarrow (x-y)^2=(x-1)^2=0\Leftrightarrow x=y=1\)