\(A=x^2-3x+1\)
\(=x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1\)
\(=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+\frac{4}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\) \(\forall\) \(x\) \(\Rightarrow\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\) \(\forall\) \(x\)
Vậy GTNN của A là \(-\frac{5}{4}\) tại \(x=\frac{3}{2}\)