\(B=3A+\dfrac{10}{A}=\dfrac{15}{\sqrt{x}+3}+10\cdot\dfrac{\sqrt{x}+3}{5}\\ B=2\left(\sqrt{x}+3\right)+\dfrac{15}{\sqrt{x}+3}\ge2\sqrt{2\left(\sqrt{x}+3\right)\cdot\dfrac{15}{\sqrt{x}+3}}\\ B\ge2\sqrt{2\cdot15}=2\sqrt{30}\\ B_{min}=2\sqrt{30}\Leftrightarrow2\left(\sqrt{x}+3\right)^2=15\\ \Leftrightarrow\sqrt{x}+3=\dfrac{\sqrt{30}}{2}\Leftrightarrow x=\dfrac{33+6\sqrt{30}}{2}\)
