\(3x^2-5x+3=3\left(x^2-\dfrac{5}{3}x+1\right)=3.\left(x^2-2.\dfrac{5}{6}x+\dfrac{25}{36}+\dfrac{11}{36}\right)=3.\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\ge\dfrac{11}{12}\)
Dấu bằng xảy ra khi
\(x-\dfrac{5}{6}=0\\ \Rightarrow x=\dfrac{5}{6}\)
Vậy \(min=\dfrac{11}{12}\) khi \(x=\dfrac{5}{6}\)