\(\left(2x+1\right)\left[\left(2x+1\right)^2-1\right]=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\2x\left(2x+2\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0;x=-1\end{matrix}\right.\)
=>(2x+1)[(2x+1)2-1]=0
=>2x(2x+1)(2x+2)=0
hay \(x\in\left\{0;-\dfrac{1}{2};-1\right\}\)
\(\left(2x+1\right)^3=2x+1\)
\(\Leftrightarrow\left(2x+1\right)\left(4x^2-2x+1\right)-\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4x^2+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4x^2+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\\x=0\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{2};-1;0\right\}.\)
