`\text{~~Holi~~}`
`(2x+1)(x^2-1)=4x^2-2x-2`
`-> (2x+1)(x^2-1)=2(2x^2-x-1)`
`-> (2x+1)(x^2-1)=2(2x^2+x-2x-1)`
`-> (2x+1)(x^2-1)=2(2x+1)(x-1)`
`-> (2x+1)(x^2-1)-2(2x+1)(x-1)=0`
`-> (2x+1)[x^2-1-2(x-1)]=0`
`-> (2x+1)(x^2-1-2x+2)=0`
`-> (2x+1)(x^2+1-2x)=0`
`->`\(\left[ \begin{array}{l}2x+1=0\\x^2+1-2x=0\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={-1/2;1}`
