Đặt \(A=9x^2-6x+2\)
\(A=9x^2-6x+1+1\\ =\left(3x-1\right)^2+1\\ \left(3x-1\right)^2\ge0\forall x\\ \Rightarrow\left(3x-1\right)^2+1\ge1\forall x\)\
Dấu "=" xảy ra khi \(\left(3x-1\right)^2=0\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)Vậy \(MIN_A=1\) khi \(x=\dfrac{1}{3}\)