Đặt \(A=-3x^2+5x+1\)
\(=-3\left(x^2-\dfrac{5}{3}x\right)+1\)
\(=-3\left[x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2\right]+3.\left(\dfrac{5}{6}\right)^2+1\)
\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{37}{12}\)
Ta thấy: \(\left(x-\dfrac{5}{6}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{37}{12}\le\dfrac{37}{12}\forall x\)
\(\Rightarrow A\le\dfrac{37}{12}\forall x\)
Dấu "=" xảy ra khi: \(x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)
Vậy GTNN của biểu thức đã cho bằng \(\dfrac{37}{12}\) khi \(x=\dfrac{5}{6}\).
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