Đặt \(A=-x^2+3x+5\)
\(=-\left(x^2-3x-5\right)\)
\(=-\left(x^2-\dfrac{3}{2}x.2+\dfrac{9}{4}-\dfrac{29}{4}\right)\)
\(=-\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{29}{4}\right]\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{29}{4}\le\dfrac{29}{4}\)
Dấu " = " khi \(-\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_A=\dfrac{29}{4}\) khi \(x=\dfrac{3}{2}\)
