\(\sqrt{x^2+4x-5}\le x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+4x-5\ge0\\x^2+4x-5\le\left(x+3\right)^2\\x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-5\\x\ge1\end{matrix}\right.\\x\ge-7\\x\ge-3\end{matrix}\right.\)
Vậy \(S=[1;+\infty)\)