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Salut437
@Salut437
May 2019
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Bonjour ! J'aimerais s'il-vous-plaît une explication pour savoir comment résoudre ces équations du second degré (factorisations) :
1) 3x(2+5x) = (2+5x)(x+1)
2) x²-4-3x(x+2)²=0
3) 3x²-12x+12=0
Merci beaucoup d'avance !
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Verified answer
Bonjour,
1) 3x(2+5x) = (2+5x)(x+1)
3x(2+5x) -(2+5x)(x+1) = 0
(2+5x)(3x-x-1) = 0
(2+5x)(2x-1)
2+5x = 0
5x = -2
x =-2/5
2x-1 =0
2x = 1
x = 1/2
2) x²-4-3x(x+2)²=0
x²-4 =(x-2)(x+2)
(x-2)(x+3)-3x(x+2)² = 0
(x+2)[(x+3)-3x(x+2)] = 0
(x+2)(x+3-3x²-6x) = 0
(x+2)(3x²-9x+3) = 0
3(x+2)(x²-3x+1) = 0
x+2 = 0
x = -2
3) 3x²-12x+12=0
avec le discri
b²-4ac = (-12)²-4(3*12) = 144-144 = 0
Δ = 0⇒1solution : -b/2a = -(-12)/6 = 2
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Verified answer
Bonjour,1) 3x(2+5x) = (2+5x)(x+1)
3x(2+5x) -(2+5x)(x+1) = 0
(2+5x)(3x-x-1) = 0
(2+5x)(2x-1)
2+5x = 0
5x = -2
x =-2/5
2x-1 =0
2x = 1
x = 1/2
2) x²-4-3x(x+2)²=0
x²-4 =(x-2)(x+2)
(x-2)(x+3)-3x(x+2)² = 0
(x+2)[(x+3)-3x(x+2)] = 0
(x+2)(x+3-3x²-6x) = 0
(x+2)(3x²-9x+3) = 0
3(x+2)(x²-3x+1) = 0
x+2 = 0
x = -2
3) 3x²-12x+12=0
avec le discri
b²-4ac = (-12)²-4(3*12) = 144-144 = 0
Δ = 0⇒1solution : -b/2a = -(-12)/6 = 2