\(-3x^2+x-20=-3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)\)
\(=-3\left(x^2-2.\dfrac{1}{6}x+\dfrac{1}{36}+\dfrac{239}{36}\right)=-3\left[\left(x-\dfrac{1}{6}\right)^2+\dfrac{239}{36}\right]\)
\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{239}{12}\le-\dfrac{239}{12}< 0\left(\forall x\right)\)
Ta có: \(-3x^2+x-20\)
\(=-3\left(x^2-\dfrac{1}{3}x+\dfrac{20}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{239}{12}\)
\(=-3\left(x-\dfrac{1}{6}\right)^2-\dfrac{239}{12}< 0\forall x\)(đpcm)