\(A=-2x^2+3x+2019\)
\(=-2\left(x^2-\frac{3}{2}x+\frac{9}{16}\right)+\frac{16161}{8}\)
\(=-2\left(x-\frac{3}{4}\right)^2+\frac{16161}{8}\) \(\le\frac{16161}{8}\)
Dấu = xảy ra \(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(Max_A=\frac{16161}{8}\Leftrightarrow x=\frac{3}{4}\)