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 \(x=\dfrac{3}{2}\)
\(-x^2+3x+5=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{29}{4}\le\dfrac{29}{4}\)Vậy giá trị lớn nhất của biểu thức là \(\dfrac{29}{4}\) khi \(-\left(x-\dfrac{3}{2}\right)^2=0\Rightarrow-x+\dfrac{3}{2}=0\Rightarrow-x=\dfrac{-3}{2}\Leftrightarrow x=\dfrac{3}{2}\)
\(-x^2+3x+5=-\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)+\dfrac{9}{4}-5=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
kl: GTLN của biểu thức là -11/4
