Bonjour, Factoriser: I(x)=(1-3x)²+(4x-1)(3-9x)+(x-2)(3x-1) J(x)=9(x-1)² - (2x+5)² Résoudre : m) 4x² + 4 x +1 = (3x-1)(2x+1) Merci d'avance (avec les étapes sv

Bonjour

I(x)=(1-3x)²+(4x-1)(3-9x)+(x-2)(3x-1)
l(x) = (1-3x)(1-3x) + 3(4x-1)(1-3x) - (x-2)(1-3x)
l(x) = (1-3x)[(1-3x)+3(4x-1)-(x-2)]
l(x) = (1-3x)(1-3x+12x-3-x+2)
l(x) = (1-3x)(8x)

J(x)=9(x-1)² - (2x+5)²
j(x) = [3(x-1)+(2x+5)][3(x-1)-(2x+5)]
j(x) = (3x-3+2x+5)(3x-3-2x-5)
j(x) = (5x+2)(x-8)

Résoudre :

m) 4x² + 4 x +1 = (3x-1)(2x+1)
4x² + 4 x +1 = (3x-1)(2x+1)
4x² + 4 x +1 = 6x²+3x-2x-1
4x² + 4 x +1 - 6x²-3x+2x+1 = 0
-2x² + 3x  + 2 = 0

Calculons Δ
Δ = b² - 4(a)(c) = 3² - 4(-2)(2) = 9 + 16 = 25
√Δ = √25 = 5

x1 = (-3 + 5) / -4 = -1/2
x2 = (-3-5) / -4 = 2

S = {-1/2;2}