\(3x^2-2x=3\left(x^2-\frac{2}{3}x\right)=3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}-\frac{1}{9}\right)\)
\(=3\left[\left(x-\frac{1}{3}\right)^2-\frac{1}{9}\right]=3\left(x-\frac{1}{3}\right)^2-\frac{1}{3}\ge\frac{-1}{3}\)
Vậy GTNN của bt là \(\frac{-1}{3}\Leftrightarrow x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)