\(B=\frac{4-4x^2+4x}{5}=\frac{-\left(4x^2-4x-4\right)}{5}\)
\(=\frac{-\left(4x^2-4x+1\right)+5}{5}\)
\(=\frac{-\left(2x-1\right)^2+5}{5}\)
Ta có: \(-\left(2x-1\right)^2\le0\)
\(\Rightarrow-\left(2x-1\right)^2+5\le5\)
\(\Rightarrow\frac{-\left(2x-1\right)^2+5}{5}\ge1\)
Vậy \(B_{min}=1\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)