6x – 3 = 0 ou 2x – 5 = 0 6x = 3 2x = 5 x x = 5 2 S

(1)

Corrigé

Exercice 1 :

(3x – 5)(2x + 4) = 0

Un produit est nul lorsque l'un des facteurs est nul.

3x – 5 = 0 ou 2x + 4 = 0 3x = 5 2x = - 4 x = 5

3 x = - 4 2 = - 2 S = { 5

3 ; - 2}

(6x – 3)(2x – 5) = 0

6x – 3 = 0 ou 2x – 5 = 0 6x = 3 2x = 5 x = 3

6 = 1

2 x = 5 2 S = { 1

2 ; 5 2 } 3x(x – 1)(2x + 7) = 0

3x = 0 ou x – 1 = 0 ou 2x + 7= 0 x = 0 x = 1 2x + 7= 0 x = - 7

2 S = { 0 ; 1 ; - 7

2}

x²(2x – 5)(- 3x +7) = 0

x² = 0 ou 2x – 5 = 0 ou - 3x +7 = 0 x = 0 2x = 5 - 3x = - 7 x = 0 x = 5

2 x = 7 3 S = { 0 ; 5

2 ; 7 3 }

Exercice 2 :

x² – 4 + 3(x – 2)(x + 7) = 0

(x – 2) (x + 2) + 3(x – 2)(x + 7) = 0 (x – 2) [(x + 2) + 3 (x + 7)] = 0 (x – 2) [ x + 2 + 3x + 21] = 0 (x – 2) (4x + 23) = 0

….

(y – 2)(y + 4) – 2(y + 4) = 0 (y + 4) [(y – 2) – 2] = 0 (y + 4) (y – 4) = 0

….

(x + 4)(x + 5) – 3(x + 4) = 0 (x + 4) [(x + 5) – 3] = 0 (x + 4) (x + 2) = 0

….

(x + 1)(2x – 5) + 5(5 – 2x) = 0 (x + 1)(2x – 5) – 5(2x – 5) = 0 (2x – 5) [(x + 1) – 5] = 0 (2x – 5) (x – 4) = 0

….

Exercice 3 :

(t – 4)(t + 2) = (t + 2)(3t – 8) (t – 4)(t + 2) – (t + 2)(3t – 8) = 0 (t + 2) [(t – 4) – (3t – 8)] = 0 (t + 2) (t – 4 – 3t + 8) = 0 (t + 2) (t – 4 – 3t + 8) = 0 (t + 2) ( - 2t + 4) = 0

t + 2 = 0 ou - 2t + 4 = 0 t = - 2 - 2t = - 4 t = - 2 t = 2

S = { 2 ; - 2}

(3x – 5)(2x + 4) = 2x + 4

(2)

- 16 + t² + 3(t – 4) = 2t – 8 t² – 16 + 3(t – 4) = 2(t – 4)

(t – 4) (t + 4) + 3(t – 4) – 2(t – 4) = 0 (t – 4) [(t + 4) + 3 – 2] = 0

(t – 4) (t + 5) = 0

t – 4 = 0 ou t + 5 = 0 t = 4 t = - 5

S = { 4 ; - 5}

(t – 5)(2t + 7) = 2t² – 5 On ne peut pas factoriser.

Exceptionnellement, on développe : 2t² + 7t – 10t – 35 = 2t² – 5

- 3t = 30 t = - 30

3 = - 10

S = { - 10 }