Les Identités Remarquables

Les Identités Remarquables

2nde – Lycée Lafayette Brioude – http://cecbertrandmath.free.fr/

( a + b ) ² =

carré 1^er nombre

a ² + 2× a × b b ²

double produit

+

+ +

(x + 3)

= x

+ 6 x + 9

Carré d’une somme

on développe

( a – b ) ² =

carré 1^er nombre

a ² – 2× a × b b ²

double produit

+

– +

Carré d’une différence

on développe

(x – 3)

= x

– 6 x + 9

(a + b) × (a – b) =

(x + 3)(x – 3) = x

– 9

Produit de la somme par la différence

carré 1^er nombre

a ² – b ²

–

on développe

Exercices 1 à 3

A(x) = (3 + x)² B(x) = (x + 5)² C(x) = (2x + 1)² D(x) = (7 + 2x)² E(x) = (3x + 2)²

EXERCICE 1 : Développer en utilisant l’identité remarquable : (a + b)² = a² + 2ab + b²

A(x) = (3 + x)² B(x) = (x + 5)² C(x) = (2x + 1)² D(x) = (7 + 2x)² E(x) = (3x + 2)²

= 9 + 6x + x² = x² + 10x + 25 = 4x² + 4x + 1 = 14 + 28x + 4x² = 9x² + 12x + 4

EXERCICE 1 : Développer en utilisant l’identité remarquable : (a + b)² = a² + 2ab + b²

A(x) = (x – 2)² B(x) = (1 – 3x)² C(x) = (3x – 2)² D(x) = (2x – 1) ² E(x) = (3 – 5x)²

= = = = =

EXERCICE 2 : Développer en utilisant l’identité remarquable : (a – b)² = a² – 2ab + b²

A(x) = (x – 2)² B(x) = (1 – 3x)² C(x) = (3x – 2)² D(x) = (2x – 1) ² E(x) = (3 – 5x)²

= x² – 4x + 4 = 1 – 6x + 9x² = 9x² – 12x + 4 = 4x² – 4x + 1 = 9 – 30x + 25x²

EXERCICE 2 : Développer en utilisant l’identité remarquable : (a – b)² = a² – 2ab + b²

A(x) = (x + 2)(x – 2) B(x) = (3x – 1)(3x + 1) C(x) = (5 + 2x )(5 – 2x) D(x) = (4 + x)(4 – x)

= = = =

EXERCICE 3 : Développer en utilisant l’identité remarquable : (a – b)(a + b) = a² – b²

A(x) = (x + 2)(x – 2) B(x) = (3x – 1)(3x + 1) C(x) = (5 + 2x )(5 – 2x) D(x) = (4 + x)(4 – x)

= x² – 4 = 9x² – 1 = 25 – 4x² = 16 – x²

EXERCICE 3 : Développer en utilisant l’identité remarquable : (a – b)(a + b) = a² – b²

( a + b ) ² = a ² + 2× a × b + b ²

(x + 3)

x

+ 6x + 9 =

on factorise

( a – b ) ² = a ² – 2× a × b + b ²

(x – 3)

x

– 6x + 9 =

on factorise

(x + 3) (x – 3) x

– 9 =

on factorise

(a + b) × (a – b) = a ² – b ²

Exercices 4 à 8

EXERCICE 4 : Factoriser en utilisant l’identité remarquable : a² + 2ab + b² = (a + b)²

A(x) = x² + 8x + 16 B(x) = x² + 12x + 36 C(x) = 4x² + 12x + 9 D(x) = 16x² + 40x + 25

= = = =

EXERCICE 4 : Factoriser en utilisant l’identité remarquable : a² + 2ab + b² = (a + b)²

A(x) = x² + 8x + 16 B(x) = x² + 12x + 36 C(x) = 4x² + 12x + 9 D(x) = 16x² + 40x + 25

= (x + 4)² = (x + 6)² = (2x + 3)² = (4x + 5)²

EXERCICE 5 : Factoriser en utilisant l’identité remarquable : a² – 2ab + b² = (a – b)²

A(x) = x² – 2x + 1 B(x) = 4x² – 20x + 25 C(x) = 9x² – 6x + 1 D(x) = 9 – 6x + x²

= = = =

EXERCICE 5 : Factoriser en utilisant l’identité remarquable : a² – 2ab + b² = (a – b)²

A(x) = x² – 2x + 1 B(x) = 4x² – 20x + 25 C(x) = 9x² – 6x + 1 D(x) = 9 – 6x + x²

= (x – 1)² = (2x – 5)² = (3x – 1)² = (3 – x)²

EXERCICE 6 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = x² – 4 B(x) = 9 – x² C(x) = x² – 16 D(x) = 4x² – 9 E(x) = 16 – 9x²

= = = = =

EXERCICE 6 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = x² – 4 B(x) = 9 – x² C(x) = x² – 16 D(x) = 4x² – 9 E(x) = 16 – 9x²

= (x + 2)(x – 2) = (3 + x)( 3 – x) = (x + 4)(x – 4) = (2x + 3)(2x – 3) = (4 – 3x)(4 + 3x)

EXERCICE 7 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = (x + 1)² – 4 B(x) = (2x + 1)² – 25 C(x) = 16 – (3x + 2)²

= = =

EXERCICE 7 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = (x + 1)² – 4 B(x) = (2x + 1)² – 25 C(x) = 16 – (3x + 2)²

= (x + 1 + 2) (x + 1 – 2)

= (x + 3) (x – 1)

= (2x + 1 + 5) (2x + 1 – 5)

= (2x + 6) (2x – 4)

= [4 + (3x + 2)] [4 – (3x + 2)]

= (4 + 3x + 2) (4 – 3x – 2)

= (3x + 6) (-3x + 2)

EXERCICE 8 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = (x + 1)² – (2x + 3)² B(x) = (2x – 1)² – (5 + x)² C(x) = (4x – 1)² – (3x – 4)²

= = =

EXERCICE 8 : Factoriser en utilisant l’identité remarquable : a² – b² = (a + b) (a – b)

A(x) = (x + 1)² – (2x + 3)² B(x) = (2x – 1)² – (5 + x)² C(x) = (4x – 1)² – (3x – 4)²

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

× [(x + 1) – (2x + 3)]

= (x + 1 + 2x + 3) (x + 1 – 2x – 3)

= (3x + 4) (-x – 2)

= [(2x – 1) + (5 + x)]

× [(2x – 1) – (5 + x)]

= (2x – 1 + 5 + x) (2x – 1 – 5 – x)

= (3x + 4) (-x – 2)

= [(4x – 1) + (3x – 4)]

× [(4x – 1) – (3x – 4)]

= (4x – 1 + 3x – 4) (4x – 1 – 3x + 4)

= (7x – 5) (x + 3)