\(A=x^2-3x+2\)
\(\Leftrightarrow A=x^2-3x+\dfrac{9}{4}-\dfrac{1}{4}\)
\(\Leftrightarrow A=\left[x^2-2.x\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]-\dfrac{1}{4}\)
\(\Leftrightarrow A=\left(x-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)
Vậy GTNN của \(A=\dfrac{-1}{4}\) khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)