Đặt \(\sqrt{4x+2}=t\ge0\Rightarrow4x+2=t^2\Rightarrow x=\frac{t^2-2}{4}\)
\(\Rightarrow B=\frac{t^2-2}{4}-5-t=\frac{t^2-4t-22}{4}=\frac{\left(t-2\right)^2-26}{4}\ge-\frac{26}{4}=-\frac{13}{2}\)
\(B_{min}=-\frac{13}{2}\) khi \(t=2\Leftrightarrow x=\frac{1}{2}\)