\(x+1=\left(x+1\right)^2\\ \Leftrightarrow x+1-\left(x+1\right)^2=0\\ \Leftrightarrow\left(x+1\right)\left(1-1-x\right)=0\\ \Leftrightarrow-x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{0;-1\right\}\)