`\sqrt{2x+1}=x-1`
$⇒\begin{cases}x-1≥0\\2x+1=(x-1)^2\end{cases}$
$⇒\begin{cases}x-1≥0\\2x+1=x^2-2x+1\end{cases}$
$⇒\begin{cases}x≥1\\x^2-4x=0\end{cases}$
$⇒\begin{cases}x≥1\\x(x-4)=0\end{cases}$
$⇒\begin{cases}x≥1\\\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\end{cases}$
`Vậy` `S={4}`