Ta có : \(-2x^2+x+5\)
= \(-2\left(x^2-\dfrac{x}{2}-\dfrac{5}{2}\right)\)
= \(-2\left(x^2-2.\dfrac{x}{4}+\dfrac{1}{4^2}-\dfrac{1}{4^2}-\dfrac{5}{2}\right)\)
= \(-2\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{41}{16}\right]\)
= \(-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{41}{8}\le\dfrac{41}{8}\) Vì \(-2\left(x-\dfrac{1}{2}\right)^2\le0\)
Vậy GTLN của đa thức là \(\dfrac{41}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)