Sagot :
Bonjour,
Rappel sur les identités remarquables ❤️ :
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
(a + b)(a - b) = a² - b²
D = 36 - 9/4x²
D = 6² - (3/2x)²
D = (6 - 3/2x)(6 + 3/2x)
H = (2 - 5x)² - (4 - 25x²)
H = (2 - 5x)(2 - 5x) - (2 - 5x)(2 + 5x)
H = (2 - 5x)(2 - 5x - (2 + 5x))
H = (2 - 5x) × 10x
H = 10x(2 - 5x)
I = 32x² + 80x + 50
I = 2(16x² + 40x + 25)
I = 2(4x + 5)²
D=36 - 9/4x^2 = 6^2 - (3/2x)^2
Du type a^2 - b^2
(6 - 3/2x)(6 + 3/2x)
H = (2 - 5x)^2 - (4 - 25x^2)
(2 - 5x)(2 - 5x) - (2^2 - (5x)^2)
(2 - 5x)(2 - 5x) - (2 - 5x)(2 + 5x)
(2 - 5x) [(2 - 5x) - (2 + 5x)]
(2 - 5x)(1 - 2 - 5x)
(2 - 5x)(-5x -1)
I = 32x^2 + 80x + 50
Delta = 80^2 -4*32*50 = 0
Une solution unique
x = -80/64 = -5/4
factorisation
32(x + 5/4)^2
Du type a^2 - b^2
(6 - 3/2x)(6 + 3/2x)
H = (2 - 5x)^2 - (4 - 25x^2)
(2 - 5x)(2 - 5x) - (2^2 - (5x)^2)
(2 - 5x)(2 - 5x) - (2 - 5x)(2 + 5x)
(2 - 5x) [(2 - 5x) - (2 + 5x)]
(2 - 5x)(1 - 2 - 5x)
(2 - 5x)(-5x -1)
I = 32x^2 + 80x + 50
Delta = 80^2 -4*32*50 = 0
Une solution unique
x = -80/64 = -5/4
factorisation
32(x + 5/4)^2