\(-3x^2+7x+5\)
\(=-3\left(x^2-\dfrac{7}{3}x-\dfrac{5}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{109}{36}\right)\)
\(=-3\left(x-\dfrac{7}{6}\right)^2+\dfrac{109}{12}< =\dfrac{109}{12}\forall x\)
Dấu '=' xảy ra khi \(x-\dfrac{7}{6}=0\)
=>\(x=\dfrac{7}{6}\)