+) Ta có:
\(P=-3x^2-x+4=-\left(3x^2+x+\dfrac{3}{36}\right)+\dfrac{49}{12}\)
\(=-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)^2+\dfrac{49}{12}\)
Vì: \(-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)^2\le0\forall x\Rightarrow-\left(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}\right)+\dfrac{49}{12}\le\dfrac{49}{12}\)
Dấu ''='' xảy ra khi \(\sqrt{3}\cdot x+\dfrac{\sqrt{3}}{6}=0\Leftrightarrow x=-\dfrac{1}{6}\)
Vậy \(MAX_P=\dfrac{49}{12}\Leftrightarrow x=-\dfrac{1}{6}\)
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