=> \(\left[{}\begin{matrix}x+9=0\\x-3=0\\2x-4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=-9\\x=3\\x=\dfrac{4}{2}=2\end{matrix}\right.\)
=>(x+9)(x-3)(x-2)=0
hay \(x\in\left\{3;2;-9\right\}\)
`12(x+9)(x-3)(2x-4)=0`
`=>` \(\left[{}\begin{matrix}x+9=0\\x-3=0\\2x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-9\\x=0+3\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=2\end{matrix}\right.\)
Vậy `S = {-9;3;2}`
`12(x+9)(x-3)(2x-4)=0`
`<=> 24(x+9)(x-3)(x-2) =0`
`<=>` \(\left[{}\begin{matrix}x+9=0\\x-3=0\\x-2=0\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=-9\\x=3\\x=2\end{matrix}\right.\)