Bonjour j’aurai besoin d’aide svp

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