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Kelyna972
@Kelyna972
May 2019
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Bjr ecq quelqu'un peux m'aider svp
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loulakar
Verified answer
Bonjour,
E(x) = (3x - 4)^2 - (2x - 5)^2
a) developper :
E(x) = 9x^2 - 24x + 16 - (4x^2 - 20x + 25)
E(x) = 5x^2 - 4x - 9
b) factoriser :
E(x) = a^2 - b^2 = (a - b)(a + b)
E(x) = (3x - 4 - 2x + 5)(3x - 4 + 2x - 5)
E(x) = (x + 1)(5x - 9)
2) résoudre :
E(x) = 0
(x + 1)(5x - 9) = 0
Pour qu’un produit soit nul il faut qu’au moins l’un de ses facteurs soit nul :
x + 1 = 0
x = -1
Ou
5x - 9 = 0
5x = 9
x = 9/5
S = {-1;9/5}
E(x) = -9
5x^2 - 4x - 9 = -9
5x^2 - 4x - 9 + 9 = 0
5x^2 - 4x = 0
x(5x - 4) = 0
x = 0
Ou
5x - 4 = 0
5x = 4
x = 4/5
S = {0;4/5}
E(x) = (2x - 5)(x + 1)
(x + 1)(5x - 9) = (2x - 5)(x + 1)
(x + 1)(5x - 9) - (x + 1)(2x - 5) = 0
(x + 1)(5x - 9 - 2x + 5) = 0
(x + 1)(3x - 4) = 0
x + 1 = 0
x = -1
Ou
3x - 4 = 0
3x = 4
x = 4/3
S = {-1;4/3}
0 votes
Thanks 1
taalbabachir
Verified answer
Soit le polynôme E(x) = (3x - 4)² - (2x - 5)² Forme 1
a) Développer , réduire et ordonner E(x) Forme 2
E(x) = (3x - 4)² - (2x - 5)²
= (9x² - 24x + 16) - (4x² - 20x + 25)
= 9x² - 24x + 16 - 4x² + 20x - 25
= 5x² - 4x - 9
identité remarquable (a - b)² = a² - 2ab + b²
b) Factoriser E(x) Forme 3
E(x) = (3x - 4)² - (2x - 5)² identité remarquable a² - b² = (a + b)(a - b)
E(x) = (3x - 4)² - (2x - 5)² = (3x - 4 + 2x - 5)(3x - 4 - 2x + 5)
= (5x - 9)(x + 1)
2. A l'aide de la forme la plus appropriée, résoudre les équations suivantes:
a) E(x) = 0 donc on utilise la forme 3 factorisée
E(x) = (5x - 9)(x + 1) = 0
5x - 9 = 0 ⇒ 5x = 9 ⇒ x = 9/5 ou
x + 1 = 0 ⇒ x = - 1
b) E(x) = - 9 ; on utilise la forme 2 développée
E(x) = 5x² - 4x - 9 = - 9 ⇔ 5x² - 4x = 0
x(5x - 4) = 0 ⇒ x = 0 ou 5x - 4 = 0 ⇒ x = 4/5
c) E(x) = (2x - 5)(x + 1) on utilise la forme 3 factorisée
E(x) = (2x - 5)(x + 1) = (5x - 9)(x + 1)
= (2x - 5)(x + 1) - (5x - 9)(x + 1) = 0
= (x + 1)(2x - 5 - 5x + 9) = 0
= (x + 1)(4 - 3x) = 0
x + 1 = 0 ⇒ x = - 1 ou 4 - 3x = 0 ⇒ x = 4/3
1 votes
Thanks 1
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Lista de comentários
Verified answer
Bonjour,E(x) = (3x - 4)^2 - (2x - 5)^2
a) developper :
E(x) = 9x^2 - 24x + 16 - (4x^2 - 20x + 25)
E(x) = 5x^2 - 4x - 9
b) factoriser :
E(x) = a^2 - b^2 = (a - b)(a + b)
E(x) = (3x - 4 - 2x + 5)(3x - 4 + 2x - 5)
E(x) = (x + 1)(5x - 9)
2) résoudre :
E(x) = 0
(x + 1)(5x - 9) = 0
Pour qu’un produit soit nul il faut qu’au moins l’un de ses facteurs soit nul :
x + 1 = 0
x = -1
Ou
5x - 9 = 0
5x = 9
x = 9/5
S = {-1;9/5}
E(x) = -9
5x^2 - 4x - 9 = -9
5x^2 - 4x - 9 + 9 = 0
5x^2 - 4x = 0
x(5x - 4) = 0
x = 0
Ou
5x - 4 = 0
5x = 4
x = 4/5
S = {0;4/5}
E(x) = (2x - 5)(x + 1)
(x + 1)(5x - 9) = (2x - 5)(x + 1)
(x + 1)(5x - 9) - (x + 1)(2x - 5) = 0
(x + 1)(5x - 9 - 2x + 5) = 0
(x + 1)(3x - 4) = 0
x + 1 = 0
x = -1
Ou
3x - 4 = 0
3x = 4
x = 4/3
S = {-1;4/3}
Verified answer
Soit le polynôme E(x) = (3x - 4)² - (2x - 5)² Forme 1a) Développer , réduire et ordonner E(x) Forme 2
E(x) = (3x - 4)² - (2x - 5)²
= (9x² - 24x + 16) - (4x² - 20x + 25)
= 9x² - 24x + 16 - 4x² + 20x - 25
= 5x² - 4x - 9
identité remarquable (a - b)² = a² - 2ab + b²
b) Factoriser E(x) Forme 3
E(x) = (3x - 4)² - (2x - 5)² identité remarquable a² - b² = (a + b)(a - b)
E(x) = (3x - 4)² - (2x - 5)² = (3x - 4 + 2x - 5)(3x - 4 - 2x + 5)
= (5x - 9)(x + 1)
2. A l'aide de la forme la plus appropriée, résoudre les équations suivantes:
a) E(x) = 0 donc on utilise la forme 3 factorisée
E(x) = (5x - 9)(x + 1) = 0
5x - 9 = 0 ⇒ 5x = 9 ⇒ x = 9/5 ou
x + 1 = 0 ⇒ x = - 1
b) E(x) = - 9 ; on utilise la forme 2 développée
E(x) = 5x² - 4x - 9 = - 9 ⇔ 5x² - 4x = 0
x(5x - 4) = 0 ⇒ x = 0 ou 5x - 4 = 0 ⇒ x = 4/5
c) E(x) = (2x - 5)(x + 1) on utilise la forme 3 factorisée
E(x) = (2x - 5)(x + 1) = (5x - 9)(x + 1)
= (2x - 5)(x + 1) - (5x - 9)(x + 1) = 0
= (x + 1)(2x - 5 - 5x + 9) = 0
= (x + 1)(4 - 3x) = 0
x + 1 = 0 ⇒ x = - 1 ou 4 - 3x = 0 ⇒ x = 4/3