Ta có: \(\left|-2x+1\right|\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
\(\Rightarrow B=\left|-2x+1\right|+3\dfrac{2}{3}\ge3\dfrac{2}{3}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(B_{min}=3\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{2}\)
