\(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\x^2=-1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(S=\left\{\dfrac{3}{2}\right\}\)