\(B=\dfrac{x^2+x}{x^2+x+1}=\dfrac{3x^2+3x}{3\left(x^2+x+1\right)}=\dfrac{-\left(x^2+x+1\right)+4x^2+4x+1}{3\left(x^2+x+1\right)}\)
\(=-\dfrac{1}{3}+\dfrac{\left(2x+1\right)^2}{3\left(x+\dfrac{1}{2}\right)^2+\dfrac{9}{4}}\ge-\dfrac{1}{3}\)
\(B_{min}=-\dfrac{1}{3}\) khi \(x=-\dfrac{1}{2}\)