Ta có : \(-3x^2+x+12\)
\(=-3\left(x^2-\dfrac{1}{3}x-4\right)\)
\(=-3\left(x^2-2x.\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{145}{36}\right)\)
\(=-3\left[\left(x-\dfrac{1}{6}\right)^2-\dfrac{145}{36}\right]\)
\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{145}{12}\le\dfrac{145}{12}\forall x\)
\(\Rightarrow\dfrac{2}{-3x^2+x+12}\ge\dfrac{2}{\dfrac{145}{12}}\)
\(\Rightarrow\dfrac{2}{-3x^2+x+12}\ge\dfrac{24}{145}\)
Dấu " = " xảy ra \(\Leftrightarrow\left(x-\dfrac{1}{6}\right)^2=0\Leftrightarrow x-\dfrac{1}{6}=0\Leftrightarrow x=\dfrac{1}{6}\)
Vậy Min của b/t trên là : \(\dfrac{24}{145}\Leftrightarrow x=\dfrac{1}{6}\)
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