a) $x^{2}$ + `13x + 12` = 0
⇔ $x^{2}$ + `12x + x + 12` = 0
⇔ `x(x + 12) + (x + 12) = 0`
⇔ `(x + 1) (x + 12) = 0`
⇔ \(\left[ \begin{array}{l}x+1=0\\x+12=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1\\x=-12\end{array} \right.\)
Vậy x ∈ {-1; -12}
b) $2x^{2}$ `-11x + 9 = 0`
⇔ $2x^{2}$ `-2x - 9x + 9 = 0`
⇔ `2x (x - 1) - 9 (x - 1) = 0`
⇔ `(2x - 9) (x - 1) = 0`
⇔ \(\left[ \begin{array}{l}2x-9=0\\x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{9}{2}\\x=1\end{array} \right.\)
Vậy x ∈ { $\dfrac{9}{2}$ , 1}
