`A=-x^{2}+5x+9`
`=-(x^{2}-5x-9)`
`=-[x^{2}-2.x.(5)/(2)+(25)/(4)-(61)/(4)]`
`=-(x-(5)/(2))^{2}+(61)/(4)\le (61)/(4)` với mọi `x`
Dấu ''='' xảy ra `<=>(x-(5)/(2))^{2}=0<=>x=5/2`
Vậy Max A `=(61)/(4)<=>x=5/2`
\(A=-\left(x^2-5x-9\right)\\ =-\left(x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}-\dfrac{61}{4}\right)\\ =-\left(x-\dfrac{5}{2}\right)^2+\dfrac{61}{4}\ge\dfrac{61}{4}\forall x\)
Dấu = xảy ra khi
\(x-\dfrac{5}{2}=0\\ x=\dfrac{5}{2}\)
Vậy \(Min_A=\dfrac{61}{4}khix=\dfrac{5}{2}\)