Ta có: \(\left(3x+1\right)^4=\left(2x-3\right)^4\)
\(\Leftrightarrow\left(3x+1\right)^4-\left(2x-3\right)^4=0\)
\(\Leftrightarrow\left[\left(3x+1\right)^2-\left(2x-3\right)^2\right]\left[\left(3x+1\right)^2+\left(2x-3\right)^2\right]=0\)
\(\Leftrightarrow\left(3x+1-2x+3\right)\left(3x+1+2x-3\right)\left(9x^2+6x+1+4x^2-12x+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(5x-2\right)\left(13x^2-6x+10\right)=0\)
mà \(13x^2-6x+10>0\forall x\)
nên \(\left[{}\begin{matrix}x+4=0\\5x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\5x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\frac{2}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-4;\frac{2}{5}\right\}\)
