Tìm x biết : $x^{2}$ + 2x – 3x – 6 = 0 $x^{2}$ + 3x – 2x – 6 = 0

a) $x^2 + 2x – 3x – 6 = 0$

⇔ $(x^2 + 2x )-( 3x + 6) = 0$

⇔ $ x(x+2) -3(x+2)=0$ 

⇔$ (x+2)(x-3)=0$ 

⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)

b) $x^2 +3x – 2x – 6 =0$ 

⇔ $(x^2 +3x )- (2x + 6)=0$ 

⇔$x (x+3) -2 (x+3) =0$ 

⇔$(x+3)(x-2)=0$ 

⇔ \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)