3(x-1)(2x-1)-5(x+8)(x-1)=0
(x-1)(3(2x-1)-5(5x+8)=0
(x-1)((6x-3)-(25x+40))=0
(x-1)(6x-3-25x-40)=0
(x-1)(-19x-43)=0
Hoặc x-1=0=>x=1
Hoặc -19x-43=0=>-19x=43=>x=-43/19
\(\left(3x-3\right)\left(2x-1\right)=\left(5x+40\right)\left(x-1\right)\)
\(6x^2-3x-6x+3=5x^2-5x+40x-40\)
\(x^2-44x+43=0\)
x=43 hoặc x=1
`3(x-1)(2x-1)=5(x+8)(x-1)`
`<=>(x-1)[3(2x-1)-5(x+8)]=0`
`<=>(x-1)(6x-3-5x-40)=0`
`<=>(x-1)(x-43)=0`
$\to \left[ \begin{array}{l}x=1\\x=43\end{array} \right.$
Vậy `S={1,43}`