Mathématiques Seconde ^= au carré Développer, réduire et ordonner : A(x)=(2x-3)(3x+4) B(x)=(x^+2x+5)(x^-x+3) C(x)=(2x+3)^ D(x)=(4x-3)^ Factoriser les expression

Bonsoir,

Developper :

A(x) = (2x - 3)(3x + 4)
A(x) = 6x^2 + 8x - 9x - 12
A(x) = 6x^2 - x - 12

B(x) = (x^2 + 2x + 5)(x^2 - x + 3)
B(x) = x^4 - x^3 + 3x^2 + 2x^3 - 2x^2 + 6x + 5x^2 - 5x + 15
B(x) = x^4 + x^3 + 6x^2 + x + 15

C(x) = (2x + 3)^2
C(x) = 4x^2 = 12x + 9

D(x) = (4x - 3)^2
D(x) = 16x^2 - 24x + 9

Factoriser :

A(x) = 2x^2 + 7x
A(x) = x(2x + 7)

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

C(x) = 4x^2 - 9
C(x) = (2x)^2 - 3^2
C(x) = (2x - 3)(2x + 3)

D(x) = (2x - 5)^2 - 16
D(x) = (2x - 5 - 4)(2x - 5 + 4)
D(x) = (2x - 9)(2x - 1)

E(x) = (3x - 1)^2 - 7
E(x) = (3x - 1 - V7)(3x - 1 + V7)

Résoudre :

1) 5x - 2 = 2x + 7
5x - 2x = 7 + 2
3x = 9
x = 9/3
x = 3

2) x/2 + 3 = 2x - 1/3
2x - x/2 = 3 + 1/3
4x/2 - x/2 = 9/3 + 1/3
3x/2 = 10/3
x = 10/3 * 2/3
x = 20/9

3) (3x + 1)(x - 2)(x + 3) = 0

Pour qu’un produit de facteur soit nul il faut qu’au moins un de ces facteur soit nul :

3x + 1 = 0
3x = -1
x = -1/3

x - 2 = 0
x = 2

x + 3 = 0
x = -3

S = {-3;-1/3;2}

4) (x + 1)^2 - 9 = 0
(x + 1 - 3)(x + 1 + 3) = 0
(x - 2)(x + 4) = 0

x - 2 = 0
x = 2

x + 4 = 0
x = -4

S = {-4;2}

5) x^2 = 5
x^2 - 5 = 0
(x - V5)(x + V5) = 0

x - V5 = 0
x = V5

x + V5 = 0
x = -V5

6) 2/x = 3/x
2/x - 3/x = 0
-1/x = 0
Avec x # 0 pas de solution

7) (3x - 1)/(x + 1) = 2
(3x - 1)/(x + 1) - 2 = 0
(3x - 1 - 2(x + 1))/(x + 1) = 0
(3x - 1 - 2x - 2)/(x + 1) = 0
(x - 3)/(x + 1) = 0

x + 1 # 0
x # -1

x - 3 = 0
x = 3