Đặt \(A=4x-x^2+3\)
\(=-x^2+4x+3=-\left(x^2-4x-3\right)\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left[\left(x-2\right)^2-7\right]\)
\(=-\left(x-2\right)^2+7\)
Ta có: \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2+7\le7\)
Dấu " = " khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)
Vậy \(MAX_A=7\) khi x = 2