Ta có: \(\left(3x+1\right)\left(x-3\right)^2=\left(3x+1\right)\left(2x-5\right)^2\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)^2-\left(3x+1\right)\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(x-3\right)^2-\left(2x-5\right)^2\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\2-x=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=2\\3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=2\\x=\frac{8}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-\frac{1}{3};2;\frac{8}{3}\right\}\)
