A=3x2-15x+2
\(=3\left(x^2-5x+\frac{3}{2}\right)\)
\(=3\left(x^2-5x+\frac{25}{4}\right)-\frac{67}{4}\)
\(=3\left(x-\frac{5}{2}\right)^2-\frac{67}{4}\ge-\frac{67}{4}\)
Dấu = khi \(x=\frac{5}{2}\)
Vậy MinA\(=-\frac{67}{4}\Leftrightarrow x=\frac{5}{2}\)
\(3x^2-15x+2=3\left(x^2-5x+\frac{25}{4}\right)-\frac{67}{4}=3\left(x-\frac{5}{2}\right)^2-\frac{67}{4}\ge0-\frac{67}{4}=-\frac{67}{4}\)
\(\Rightarrow MIN_{3x^2-15x+2}=-\frac{67}{4}\Leftrightarrow x=\frac{5}{2}\)