Bonjour aidez moi svp pour la c) et la d) merci d'avance !!!

Bonjour Yasmine.

c/ 2 methodes possibles :
(4x - 3)(2x/3 - 1) = 0
1re methode :
8x²/3 - 4x - 2x + 3 = 0
(8/3)x² - 6x + 3 = 0
Δ = 36 - 4 * 8/3 * 3
Δ = 36 - 32
Δ = 4
√Δ = 2
Δ > 0 ⇒ 2 solutions :
x₁ = (6 - 2) / 2 * 8/3 = 4 / (16/3) = 3/4
x₂ = (6 + 2) / 2 * 8/3 = 8 / (16/3) = 3/2
S = {3/4 ; 3/2}
2e methode :
2 solutions :
4x - 3 = 0 ⇔ 4x = 3 ⇔ x = 3/4
(2/3)x - 1 = 0 ⇔ (2/3)x = 1 ⇔ x = 3/2
S = {3/4 ; 3/2}

d/ 2 methodes possibles egalement.
2(3x - 8)(x/2 + 1) = 0
2(x/2 + 1)(3x - 8) = 0
(x + 2)(3x - 8) = 0
2 solutions :
x + 2 = 0 ⇔ x = - 2
3x - 8 = 0 ⇔ 3x = 8 ⇔ x = 8/3
S = {- 2 ; 8/3}

Bonne semaine !