\(C=5x-x^2\)
\(=-\left(x^2-5x+\dfrac{25}{4}\right)+\dfrac{25}{4}\)
\(=-\left[x^2-2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\dfrac{25}{4}\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Ta có : \(-\left(x-\dfrac{5}{2}\right)^2\le0\Rightarrow-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
Dấu = xảy ra \(\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(Max_C=\dfrac{25}{4}\Leftrightarrow x=\dfrac{5}{2}\)
\(D=-x^2+6x-11\)
\(=-\left(x^2-6x+11\right)\)
\(=-\left[\left(x^2-6x+9\right)+2\right]\)
\(=-\left(x-3\right)^2-2\)
Ta có :\(-\left(x-3\right)^2\le0\Leftrightarrow-\left(x-3\right)^2-2\le-2\)
Dấu = xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(Max_D=-2\Leftrightarrow x=3\)
