\(A=x^2-3x+2\\ \Rightarrow A=\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{1}{4}\\ \Rightarrow A=\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(A_{min}=-\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{2}\)