\(A=\dfrac{x+1}{x-3}=\dfrac{x-3+4}{x-3}=1+\dfrac{4}{x-3}\)
\(\left|A\right|=3=\left|1+\dfrac{4}{x-3}\right|\)
\(\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{4}{x-3}=3\\1+\dfrac{4}{x-3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-3}=2\\\dfrac{4}{x-3}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)