Ta có: \(B-\dfrac{2}{3}=\dfrac{x^2+1}{x^2-x+1}-\dfrac{2}{3}=\dfrac{\left(x-1\right)^2}{3\left(x^2-x+1\right)}=\dfrac{\left(x-1\right)^2}{3\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}\ge0\Rightarrow B\ge\dfrac{2}{3}\).
Đẳng thức xảy ra khi x = 1.
Vậy Min B = \(\frac{2}{3}\) khi x = 1.