Đặt \( (x-1)^2=t\)
\(→t^2-3t-4=0\\↔t^2-4t+t-4=0\\↔(t^2-4t)+(t-4)=0\\↔t(t-4)+(t-4)=0\\↔(t+1)(t-4)=0\\↔\left[\begin{array}{1}t+1=0\\t-4=0\end{array}\right.\\↔\left[\begin{array}{1}t=-1\\t=4\end{array}\right.\\↔\left[\begin{array}{1}(x-1)^2=-1(vô\,\,lý)\\(x-1)^2=4\end{array}\right. \\↔\left[\begin{array}{1}x-1=2\\x-1=-2\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=-1\end{array}\right.\)
Vậy \(S=\{3;-1\}\)
