Svp aidez moi pour mon dm!C'est des equations

Bonsoir,

(3x - 6)(x + 5) - (3x + 1)(2x - 4) = 0
3(x - 2)(x + 5) - 2(3x + 1)(x - 2) = 0
(x - 2)[3(x + 5) - 2(3x + 1)] = 0
(x - 2)(3x + 15 - 6x - 2) = 0
(x - 2)(-3x + 13) = 0

Pour qu’un produit soit nul il faut qu’un de ses facteurs soit nul :

x - 2 = 0
x = 2
Ou
-3x + 13 = 0
3x = 13
x = 13/3

(3x + 2)(x - 5) - (10 - 2x)(x + 4) + (15 - 3x)(x - 3) = 0
(3x + 2)(x - 5) + 2(-5 + x)(x + 4) - 3(-5 + x)(x - 3) = 0
(x - 5)[3x + 2 + 2(x + 4) - 3(x - 3)] = 0
(x - 5)(3x + 2 + 2x + 8 - 3x + 9) = 0
(x - 5)(2x + 19) = 0

x - 5 = 0
x = 5
Ou
2x + 19 = 0
2x = -19
x = -19/2

(7 - 3x)(x + 4) - 2(3x - 7) - 7 + 3x = 0
(7 - 3x)(x + 4) + 2(7 - 3x) - (7 - 3x) = 0
(7 - 3x)(x + 4 + 2 - 1) = 0
(7 - 3x)(x + 5) = 0

7 - 3x = 0
3x = 7
x = 7/3
Ou
x + 5 = 0
x = (-5)