Ta có:
(3x-1)(x²+2) = (3x-1)(7x-10)
⇔ (3x-1)(x²+2) - (3x-1)(7x-10) = 0
⇔ (3x-1)(x²+2-7x+10) = 0
⇔ (3x-1)( x² - 7x + 12 ) = 0
⇔ (3x-1)( x² - 3x - 4x + 12 ) = 0
⇔ (3x-1)[x(x-3) - 4(x-3) ] = 0
⇔ (3x-1)(x-3)(x-4) = 0
⇔ \(\left[ \begin{array}{l}3x - 1=0\\x-3=0 \end{array} \right.\) \
hoặc x - 4= 0
⇔ \(\left[ \begin{array}{l}x= \frac{1}{3}\\x=3\end{array} \right.\)
hoặc x = 4
Vậy S = { $\frac{1}{3}$ , 3, 4 }