\(A=-4x^2+12x-2\)
\(=-\left(4x^2-12x+2\right)\)
\(=-\left[\left(2x\right)^2-2.2x.3+3^2-3^2+2\right]\)
\(=-\left[\left(2x-3\right)^2-7\right]\)
\(=-\left(2x-3\right)^2+7\)
Ta có \(\left(2x-3\right)^2\ge0\)
\(\Leftrightarrow-\left(2x-3\right)^2\le0\)
\(\Leftrightarrow-\left(2x-3\right)^2+7\le7\)
Vậy MAX của \(A=1\Leftrightarrow-\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)