\(B=\dfrac{x+8}{\sqrt{x}-1}=\dfrac{x-1}{\sqrt{x}-1}+\dfrac{9}{\sqrt{x}-1}\)
\(=\sqrt{x}+1+\dfrac{9}{\sqrt{x}-1}\)
\(=\sqrt{x}-1+\dfrac{9}{\sqrt{x}-1}+2\)
\(\ge2\sqrt{\left(\sqrt{x}-1\right)\dfrac{9}{\sqrt{x}-1}}+2=8\)
\(MinB=8\Leftrightarrow\sqrt{x}-1=\dfrac{9}{\sqrt{x}-1}\)
\(\Leftrightarrow x=16\)

