a) x + 5x² = 0
⇔ x.(1 + 5x) = 0
⇔ \(\left[ \begin{array}{l}x = 0\\1 + 5x =0 \end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 0\\x = -\frac{1}{5}\end{array} \right.\)
Vậy x ∈ {0; -$\frac{1}{5}$}
b) (x + 1) = (x + 1)²
⇔ (x + 1) - (x + 1)² = 0
⇔ (x + 1)(1 - x - 1) = 0
⇔ (x + 1).(-x) = 0
⇔ \(\left[ \begin{array}{l}x + 1 = 0\\-x = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = -1\\x = 0\end{array} \right.\)
Vậy x ∈ {0; -1}
c) x³ + x = 0
⇔ x.(x² + 1) = 0
⇔ \(\left[ \begin{array}{l}x = 0\\x² + 1 = 0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x = 0\\x² = -1 (Vô lí)\end{array} \right.\)
⇔ x = 0
Vậy x = 0
