\(B=\frac{x}{x^2+4}-\frac{3}{13}+\frac{3}{13}=\frac{13x-3x^2-12}{x^2+4}+\frac{3}{13}=\frac{\left(3-x\right)\left(3x-4\right)}{x^2+4}+\frac{3}{13}\)
Do \(x\ge3\Rightarrow\left\{{}\begin{matrix}3-x\le4\\3x-4>0\end{matrix}\right.\) \(\Rightarrow\frac{\left(3-x\right)\left(3x-4\right)}{x^2+4}\le0\)
\(\Rightarrow B\le\frac{3}{13}\Rightarrow B_{max}=\frac{3}{13}\) khi \(x=3\)