Bonjour,
Exercice 4 :
(x - 2)(x - 1) + (x - 2)(2x + 4) = 0
(x - 2)(x - 1 + 2x + 4) = 0
(x - 2)(3x + 3) = 0
3(x - 2)(x + 1) = 0
Un produit est nul si au moins l’un de ses facteurs est nul :
3 # 0
Donc :
x - 2 = 0
x = 2
Ou
x + 1 = 0
x = -1
S = {-1;2}
3(x + 5) - 2x = 7
3x + 15 - 2x = 7
x = 7 - 15
x = -8
x^2 + x = 2x^2
2x^2 - x^2 - x = 0
x^2 - x = 0
x(x - 1) = 0
x = 0
Ou
x - 1 = 0
x = 1
x^2 + 1 = 2x
x^2 - 2x + 1 = 0
x^2 - 2 * x * 1 + 1^2 = 0
(x - 1)^2 = 0
x - 1 = 0
x = 1
Exercice 6 :
1) f(x) = 1
x = -2,5 ; x = 3 et x = 6,5
2) f(x) = 0
x = -2 ; x = 2 et x = 7
3) g(x) = -3
x = 8
4) f(x) = g(x)
x = -3 ; x = 4 et x = 6
Exercice 2 :
(2x - 4)(x + 5) + x > -5
(2x - 4)(x + 5) + x + 5 > 0
(x + 5)(2x - 4 + 1) > 0
(x + 5)(2x - 3) > 0
(2x - 3)(x + 5) > 0 vrai l’egalite Est prouvée
Résoudre :
(2x - 4)(x + 5) + x > -5
(2x - 3)(x + 5) > 0
2x - 3 > 0
2x > 3
x > 3/2
x + 5 < 0
x < -5
x..............|..-inf.............(-5)...........3/2.........+inf
2x - 3.....|...........(-)................(-)......O....(+).........
x + 5......|...........(-).........O......(+).............(+).........
Eq..........|...........(+)........O......(-).....O....(+)..........
Eq > 0 pour x € ]-inf ; -5[ U ]3/2 ; + inf [
