Ta có: \(A=9x^2-6x+\dfrac{7}{4}\)
\(=\left(3x\right)^2-2.3x+1+\dfrac{3}{4}\)
\(=\left(3x-1\right)^2+\dfrac{3}{4}\)
Ta lại có: \(\left(3x-1\right)^2\ge0\)
\(\Rightarrow\left(3x-1\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Rightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra \(\Leftrightarrow\left(3x-1\right)^2=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy MinA = \(\dfrac{3}{4}\) \(\Leftrightarrow x=\dfrac{1}{3}\)
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