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Crystaliaxafira
@Crystaliaxafira
May 2019
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Bonjour besoin d'aide pour les équations / factorisations en 3ème svp.
1) Développer et réduire : A = (2x+1)2 (au carré) -3(2x+1)(x-3)
2) Factoriser A.
3) Résoudre l'équation : (2x+1)(-x+10) = 0.
Merci d'avance.
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alphis
Pour la 3 il te suffit de faire :
(2x+1)(-x+10) = 0
2x+1=0 ou -x+10=0
2x+1-1 =-1 / -x+10+x=+x
x=-0.5 / 10 = x
0 votes
Thanks 1
Christine17
A = (2x+1)²-3(2x+1)(x-3)
A=(2x+1)((2x+1)-3(x-3))
A=(2x+1)(2x+1-3x+9)
A=(2x+1)(-x+10)
3) Résoudre l'équation : (2x+1)(-x+10) = 0.
2x+1=0
0U -x+10= 0.
x=-1/2 ou x=10
0 votes
Thanks 0
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2x+1=0 ou -x+10=0
2x+1-1 =-1 / -x+10+x=+x
x=-0.5 / 10 = x
A=(2x+1)((2x+1)-3(x-3))
A=(2x+1)(2x+1-3x+9)
A=(2x+1)(-x+10)
3) Résoudre l'équation : (2x+1)(-x+10) = 0.
2x+1=0
0U -x+10= 0.
x=-1/2 ou x=10