Réponse:

pour l'exercice 29, le A, c est une identité remarquable

a= ( x - 2)²

a= x² - 2 × x × 2 +2²

a= x² - 4x + 4

b= pour celle ci c est une double distributivité

b= x × x + x × (-1) + 3 × x + 3 × (-1)

b= x² - 1x + 3x + 3

b= x² + 2x + 3

et tu applique ca pour tout l exercice 29

J ESPÈRE QUE JE T AURAI UN PEU AIDER

Bonjour !

Exercice 29 :

A = (x - 2)² = x² - 4x + 4

B = (x + 3)(x - 1) = x² + 3x - x - 3 = x² + 2x - 3

C = (x - 4)(2 - x) = 2x - 8 - x² + 4x = 6x - 8 - x²

D = (x - 2)(x + 2) = x² - 4

Exercice 30 :

A = (x + 2)(3x - 1) = 3x² + 6x - x - 2 = 3x² + 5x - 2

B = (2x - 1)² = 4x² - 4x + 1

C = (4x + 1)² = 16x² + 8x + 1

D = (2x - 1)(-3x + 1) = -6x² + 3x + 2x - 1 = -6x² + 5x - 1

Exercice 31 :

A = 3(x - 2)(x + 5) = 3(x² - 2x + 5x + 3) = 3x² - 6x + 15x + 9 = 3x² + 9x + 9

B = -2(x + 3)(2x - 4) = -2(2x² + 6x - 4x - 12) = -4x² - 12x + 8x + 24

= -4x²-4x + 24

C = (2 - x²)(3 + x) = 6 - 3x² + 2x - x³

D = 2(x - 3)² = 2(x² - 6x + 9) = 2x² - 12x + 18

Exercice 32 :

A = (x + 1)(x - 4) + (x + 3)(x - 1)

A = x² + x - 4x - 4 + x² + 3x - x - 3

A = 2x² - x - 7

B = (x + 1)² - (x + 3)(x - 5)

B = x² + 2x + 1 - (x² + 3x - 5x - 15)

B = x² + 2x + 1 - x² - 3x + 5x + 15

B = 4x + 16

Exercice 38 :

A = (x - 5)(x + 2) + (x - 5)(x - 7)

A = (x - 5)[(x + 2) + (x - 7)]

A = (x - 5)(x + 2 + x - 7)

A = (x - 5)(2x - 5)

B = x² - 10x

B = x(x - 10)

Exercice 39 :

A = x(x + 2) + 3x

A = x(x + 2 + 3)

A = x(x + 5)

B = x² - 5x

B = x(x - 5)

C = (x + 2)(x + 1) - 2(x + 1)

C = (x + 1)(x + 2 - 2)

C = x(x  + 1)

D = (x + 3)(x - 2) - (x - 2)(2x + 1)

D = (x - 2)[(x + 3) - (2x + 1)]

D = (x - 2)(x + 3 - 2x - 1)

D = (x - 2)(2 - x)

J'espère t'avoir aidé. Bon courage !