\(A=B\cdot\left|\sqrt{x}-3\right|\)
\(\Leftrightarrow\left|\sqrt{x}-3\right|=\dfrac{x-3}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}=x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(x-3-\sqrt{x}+3\right)\left(x-3+\sqrt{x}-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(x-\sqrt{x}\right)\left(x+\sqrt{x}-6\right)=0\end{matrix}\right.\Leftrightarrow x=4\)