\(B=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}\)
Ta có \(x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(\Leftrightarrow B\le2+\dfrac{1}{\dfrac{3}{4}}=2+\dfrac{4}{3}=\dfrac{10}{3}\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy \(B_{max}=\dfrac{10}{3}\)