\(\sqrt{x}+3\ge3\Rightarrow\dfrac{5}{\sqrt{x}+3}\le\dfrac{5}{3}\)
\(A_{max}=\dfrac{5}{3}\) khi \(x=0\)
\(B=3A+\dfrac{10}{A}=3\left(A+\dfrac{25}{9A}\right)+\dfrac{5}{3A}\ge3.2\sqrt{\dfrac{25A}{9A}}+\dfrac{5}{3.\dfrac{5}{3}}=11\)
\(B_{min}=11\) khi \(A=\dfrac{5}{3}\)