Vì \(x^2-2x+4>0\forall x\)
nên \(x^2-2x+4=2x+1\)
\(\Leftrightarrow x^2-2x+4-2x-1=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy: S={1;3}