ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)
Ta có: \(12\sqrt{x+1}-x-1=0\)
\(\Leftrightarrow12\sqrt{x+1}=x+1\)
\(\Leftrightarrow a^2=12a\)
\(\Leftrightarrow a^2-12a=0\)
\(\Leftrightarrow a\left(a-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{x+1}=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+1=144\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=143\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-1;143\right\}\)
