\(N=2x-2x^2-5=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le\dfrac{-9}{2}\)
Dấu " = " khi \(-2\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(MAX_N=\dfrac{-9}{2}\) khi \(x=\dfrac{1}{2}\)