\(B=\dfrac{2}{3}+\dfrac{\left(2x-1\right)^2}{x^2+2x+3}=\dfrac{2}{3}+\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\)
vì \(\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\ge0\)
\(B_{nn}\Leftrightarrow\dfrac{\left(2x-1\right)^2}{\left(x+1\right)^2+2}\)(nn)
\(\Rightarrow B\ge\dfrac{2}{3}\)
B(nn)=\(\dfrac{2}{3}\) ; khi 2x-1 =0 hay x=1/2