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Pomme22
@Pomme22
May 2019
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Bonjour,
J'ai un petit probleme en math il faut que je resoude f(x)=0 et f(x)=-2
pour f=(3x-2)(x+4)-(3x-2)(5x+3)
j'ai developpé et j'ai trouve -13 x au carré +9 x - 14 comment doit je m'y prendre ?
Merci
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Verified answer
Bonjour,
f(x) = 0 ne pas developper, factoriser
(3x-2)(x+4)-(3x-2)(5x+3) = 0
(3x-2)(x+4-5x-3) = 0
(3x-2)(-4x+1) = 0
3x-2 = 0
3x =2
x = 2/3
-4x+1 = 0
-4x =-1
x = 1/4
f(x) = -2
on developpe = -12x²+11x-2
-12x²+11x-2 =-2
-12x²+11x-2+2 = 0
-12x²+11x = 0
x(-12x+11) = 0
-12x+11 =0
-12x =-1
x =11/12
x = 0
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Thanks 1
Pomme22
Merci j'ai compris le premier mais pourquoi dans le 2 eme calcul le x(-12x+11)=0 se transforme en -12x+11=0 ligne suivante le x à disparu ?
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Verified answer
Bonjour,f(x) = 0 ne pas developper, factoriser
(3x-2)(x+4)-(3x-2)(5x+3) = 0
(3x-2)(x+4-5x-3) = 0
(3x-2)(-4x+1) = 0
3x-2 = 0
3x =2
x = 2/3
-4x+1 = 0
-4x =-1
x = 1/4
f(x) = -2
on developpe = -12x²+11x-2
-12x²+11x-2 =-2
-12x²+11x-2+2 = 0
-12x²+11x = 0
x(-12x+11) = 0
-12x+11 =0
-12x =-1
x =11/12
x = 0