\(ĐKXĐ:x\ge1\)
\(\sqrt{x^2+2x+2}=x-1\)
\(\Rightarrow\left(\sqrt{x^2+2x+2}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow\left|x^2+2x+2\right|=x^2-2x+1\)
\(\Leftrightarrow x^2+2x+2=x^2-2x+1\) (do \(x^2+2x+2>0\forall x\in R\)).
\(\Leftrightarrow4x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{4}\left(loại\right)\)
Vậy S=∅