Réponse :
a) 2/(3 x + 1) = 5 V.I : x ≠ - 1/3
⇔ 2/(3 x + 1) - 5 = 0
⇔ 2/(3 x+1) - 5(3 x+1)/(3 x+1) = 0
⇔ (2 - 15 x - 5)/(3 x + 1) = 0
⇔ (- 15 x - 3)/(3 x + 1) = 0
⇔ - 15 x - 3 = 0
⇔ x = - 3/15 = - 1/5
or - 1/5 ≠ - 1/3 donc S ={- 1/5}
c) (3 x + 1)/(6 - 5 x) = 2 : V.I : x ≠ 6/5
⇔ (3 x + 1)/(6 - 5 x) - 2 = 0
⇔ (3 x + 1)/(6 - 5 x) - 2(6 - 5 x)/(6 - 5 x) = 0
⇔ (3 x + 1 - 12 + 10 x)/(6 - 5 x) = 0
⇔ (13 x - 11)/(6 - 5 x) = 0
⇔ 13 x - 11 = 0
⇔ x = 11/13
or 11/13 ≠ 6/5 donc S = {11/13}
e) 3/(x- 1) = 4/(1 - 2 x) : V.I x ≠ 1 et x ≠ 1/2
⇔ 3/(x- 1) - 4/(1 - 2 x) = 0
⇔ 3(1 - 2 x)/(x-1)(1 - 2 x) - 4(x - 1)/(x-1)(1-2 x) = 0
⇔ (3 - 6 x - 4 x + 4)/(x-1)(1 - 2 x) = 0
⇔ (- 10 x + 7)/(x-1)(1-2 x) = 0
⇔ - 10 x + 7 = 0
⇔ x = 7/10
or 7/10 ≠ 1 et 7/10 ≠ 1/2 donc S = {7/10}
Explications étape par étape
