3(2-x)+5(x-6)=-98
⇒ 6x -3 +5x -30 = -98
⇒ 11x -33 = -98
⇒ 11x = -65
⇒ x = $\frac{-65}{11}$
Vậy x = $\frac{-65}{11}$
(x²-36)(x³+27)=0
⇒ \(\left[ \begin{array}{l}x²-36=0\\x³+27=0\end{array} \right.\)
⇒ \(\left[ \begin{array}{l}x=±6\\x=-3\end{array} \right.\)
Vậy x ∈ {-6; 6; -3}