2/
\(x^3-2x+1=0\)
\(\Rightarrow x^3-x-x+1=0\)
\(\Rightarrow x\left(x^2-1\right)-\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left(x+1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+x-1\right)\)
\(\Rightarrow x=1\)
Vậy S = {1}