Tìm x biết : $x^{2}$ + 2x – 3x – 6 = 0 $x^{2}$ + 3x – 2x – 6 = 0 01/07/2021 Bởi Natalia 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) $x^2 +3x – 2x – 6 =0$ ⇔ $(x^2 +3x )- (2x + 6)=0$ ⇔$x (x+3) -2 (x+3) =0$ ⇔$(x+3)(x-2)=0$ ⇔ \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\) Bình luận
Bạn tham khảo: $x^{2}+2x-3x-6=0$ ⇔$x(x+2)-3(x-2)=0$ ⇔$(x+2).(x-3)=0$ ⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\) $x^{2}+3x-2x-6=0$ ⇔$x(x+3)-2(x+3)=0$ ⇔$(x+3).(x-2)=0$ ⇔\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) HỌC TỐT@Chin…. Bình luận
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) $x^2 +3x – 2x – 6 =0$
⇔ $(x^2 +3x )- (2x + 6)=0$
⇔$x (x+3) -2 (x+3) =0$
⇔$(x+3)(x-2)=0$
⇔ \(\left[ \begin{array}{l}x=-3\\x=2\end{array} \right.\)
Bạn tham khảo:
$x^{2}+2x-3x-6=0$
⇔$x(x+2)-3(x-2)=0$
⇔$(x+2).(x-3)=0$
⇔\(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
$x^{2}+3x-2x-6=0$
⇔$x(x+3)-2(x+3)=0$
⇔$(x+3).(x-2)=0$
⇔\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
HỌC TỐT
@Chin….