\(B=\frac{6\sqrt{x}-3}{3\left(x+2\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1-x+4\sqrt{x}-4}{3\left(x+2\sqrt{x}+1\right)}=\frac{1}{3}-\frac{\left(\sqrt{x}-2\right)^2}{3\left(\sqrt{x}+1\right)^2}\le\frac{1}{3}\)
\(B_{max}=\frac{1}{3}\) khi \(\sqrt{x}=2\Leftrightarrow x=4\)