$\color{pink}{\text{sakurangoc2007}}$
Xin hay nhất nha!! 🧡
a) 3x(x-1)+x-1=0
⇔ 3x (x - 1) + (x - 1) = 0
⇔ (3x + 1) (x - 1) = 0
⇔ \(\left[ \begin{array}{l}3x+1=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}3x=-1\\x=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=1\end{array} \right.\)
Vậy x ∈ {$\dfrac{-1}{3}$ ; 1}
b) 2 (x + 3)- $x^2$ - 3x = 0
⇔ 2 (x + 3) - x (x + 3) = 0
⇔ (2 - x) (x + 3) = 0
⇔ \(\left[ \begin{array}{l}2 - x=0\\x+3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
Vậy x ∈ {2; -3}
