Ta có:\(2x-2x^2-5=-\left(2x^2-2x+5\right)\)
\(=-\left[2\left(x^2-x+\dfrac{5}{2}\right)\right]\)
\(=-\left\{2\left[x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+\dfrac{5}{2}\right]\right\}\)
\(=-\left\{2\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\right]\right\}\)
\(=-\left[2\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{2}\right]\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)
Do \(-2\left(x-\dfrac{1}{2}\right)^2\le0\) với \(\forall x\) (dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\) )
\(\Rightarrow-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\) hay \(2x-2x^2-5\le-\dfrac{9}{2}\) (dấu ''='' xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\))
Vậy giá trị lớn nhất của biểu thức \(2x-2x^2-5\) là \(-\dfrac{9}{2}\) tại \(x=\dfrac{1}{2}\)
