\(A=1-\left|3x-1\right|+\left(3x-1\right)^2\)
Đặt \(\left|3x-1\right|=a\ge0\)
\(A=a^2-a+1=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow A_{min}=\frac{3}{4}\) khi \(a=\frac{1}{2}\Leftrightarrow\left|3x-1\right|=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{1}{6}\end{matrix}\right.\)