\(B=3x^2-3x+1\)
\(B=3\left(x^2-x+\frac{1}{3}\right)\)
\(B=3\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)\)
\(B=3\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)
Vậy \(B_{min}=\frac{1}{4}\Leftrightarrow x=\frac{1}{2}\)