Bonjour,
1)
Identité remarquable: (a + b)(a - b) = a² - b²
A = (6 + 5x)(6 - 5x) = 6² - (5x)² = 36 - 25x² = A
2)
B = (x + 9)(5 - x) - 7(x + 9)(4x + 3) = (x + 9)(5 - x - 7(4x + 3))
B = (x + 9)(5 - x - 28x - 21) = (x + 9)(-29x - 16) = B
C = 64x² - 100 = (8x)² - 10² = (8x + 10)(8x - 10) = C (Identité remarquable vu en question 1)
3)
a)
D(-1) = 36 - (1 - 2 * (-1)) = 36 - (1 + 2) = 36 - 3 = 33 = D(-1)
b)
D = 36 - (1 - 2x) = 36 - 1 + 2x = 35 + 2x = D
c)
D = 36 - (1 - 2x) = 35 + 2x Non factorisable.
Si tu as par hasard oublié un carré:
D = 36 - (1 - 2x)² = 6² - (1 - 2x)² = (6 - (1 - 2x))(6 + (1 - 2x))
D = (6 - 1 + 2x)(6 + 1 - 2x) = (5 + 2x)(7 - 2x) = D
Dans ce cas:
D(-1) = 27 (Q3a) et D = 36 - 1 + 4x - (2x)² = 35 + 4x - 4x² = D (Q3b)
4)
(4x - 5)(18 + 3x) = 0
4x - 5 = 0 ou 18 + 3x = 0
<=> 4x = 5 ou 3x = -18
<=> x = 5/4 ou x = -6
Bonne journée,
Thomas
