a) \(A=5-3.\left(3x-1\right)^2=-\left[3\left(3x-1\right)^2-5\right]\)
Ta có: \(\left(3x-1\right)^2\ge0\forall x\)
\(\Rightarrow3.\left(3x-1\right)^2\ge0\)
\(\Rightarrow3\left(3x-1\right)^2-5\ge-5\forall x\)
\(\Rightarrow-\left[3\left(3x-1\right)^2-5\right]\ge5\forall x\)
Vậy \(MinA=5\Leftrightarrow x=\dfrac{1}{3}\)