a) (x - 8) (x³ + 8) = 0
⇔ \(\left[ \begin{array}{l}x-8=0\\x³+ 8=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=8\\x³=-8\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=8\\x³=(-2)³\end{array} \right.\)
⇔ ⇔ \(\left[ \begin{array}{l}x=8\\x=-2\end{array} \right.\)
Vậy x ∈ {8 ; -2}
b) (4x - 3) – ( x + 5) = 3(10 - x)
⇔ 4x - 3 - x - 5 = 30 - 3x
⇔ (4x - x) - (3 +5) - 30 + 3x = 0
⇔ 3x - 8 - 30 + 3x = 0
⇔ 6x - 38 = 0
⇔ 6x = 38
⇔ x = $\frac{38}{6}$
⇔ x = $\frac{19}{3}$
Vậy x = $\frac{19}{3}$
