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Andrea35Miller
@Andrea35Miller
April 2019
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SVP aidez moi
B(x) =(3x+4)(5x+2)-(3x+4)(2x-5)
1.developper et réduire
2.Factoriser B(x)
3.Resoudre l'équation B(x)=0
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queenomega
1)
B(x) =(3x+4)(5x+2)-(3x+4)(2x-5)
3x*5x+3x*2+4*5x+4*2-3x*2x-3x*(-5)-4*2x-4*(-5)
15x²+6x+20x+8-6x²+15x-8x+20
9x²++33x+28
2)
(3x+4)(3x+7)
on développe pour vérifier que notre factorisation et notre developpement sont bons
3x*3x+3x*7+4*3x+4*7
9x²+21x+12x+28
9x²+33x+28
c est le cas
3)
B(x)=0
(3x+4)(5x+2)-(3x+4)(2x-5)=(3x+4)(3x+7)
3x+4=0 ou 3x+7=0
3x=-4 3x=-7
x=-4/3 x=-7/3
d'ou S= {-4/3 ; -7/3 }
1 votes
Thanks 2
Eliott78
Je trouve la même chose que Queenomega !!
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Lista de comentários
B(x) =(3x+4)(5x+2)-(3x+4)(2x-5)
3x*5x+3x*2+4*5x+4*2-3x*2x-3x*(-5)-4*2x-4*(-5)
15x²+6x+20x+8-6x²+15x-8x+20
9x²++33x+28
2)
(3x+4)(3x+7)
on développe pour vérifier que notre factorisation et notre developpement sont bons
3x*3x+3x*7+4*3x+4*7
9x²+21x+12x+28
9x²+33x+28
c est le cas
3)
B(x)=0
(3x+4)(5x+2)-(3x+4)(2x-5)=(3x+4)(3x+7)
3x+4=0 ou 3x+7=0
3x=-4 3x=-7
x=-4/3 x=-7/3
d'ou S= {-4/3 ; -7/3 }