\(2A=-10x^2-4y^2+4xy+8x-4y+4038\)
\(=-\left(4x^2+y^2+1-4xy-4x+2y\right)-\dfrac{2}{3}\left(9x^2-6x+1\right)-\dfrac{1}{3}\left(9y^2+6y+1\right)+4040\)
\(=-\left(2x-y-1\right)^2-\dfrac{2}{3}\left(3x-1\right)^2-\dfrac{1}{3}\left(3y+1\right)^2+4040\le4040\)
\(\Rightarrow A\le2020\)
\(A_{max}2=2020\) khi \(\left(x;y\right)=\left(\dfrac{1}{3};-\dfrac{1}{3}\right)\)