a) 6x² - 12x = 0
⇔ 6x(x - 2) = 0
⇔ \(\left[ \begin{array}{l}6x = 0\\x - 2 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 0\\x = 2\end{array} \right.\)
Vậy x ∈ {0; 2}
b) (x + 1) = (x + 1)³
⇔ (x + 1) - (x + 1)³ = 0
⇔ (x + 1)[1 - (x + 1)²] = 0
⇔ (x + 1)(1 - x² - 2x - 1) = 0
⇔ (x + 1)(-x² - 2x) = 0
⇔ (x + 1).(-x).(x + 2) = 0
TH1: x + 1 = 0
⇔ x = -1
TH2: -x = 0
⇔ x = 0
TH3: x + 2 = 0
⇔ x = -2
Vậy x ∈ {-2; -1; 0}
c) 3x(x - 2020) - x +2020 = 0
⇔ 3x(x - 2020) - (x - 2020) = 0
⇔ (x - 2020)(3x - 1) = 0
⇔ \(\left[ \begin{array}{l}x - 2020 = 0\\3x - 1 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 2020\\x = \frac{1}{3}\end{array} \right.\)
Vậy x ∈ {2020; $\frac{1}{3}$}
