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Andrea35Miller
@Andrea35Miller
April 2019
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Bonjour si vous pourriez m'aider pour le petit 2 et le petit 5 ça serait cool thxx
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esefiha
2) factoriser B(x)
B(x) = (3x+4)(5x+2)-(3x+4)(2x-5)
on met 3x+4 en facteur
B(x) = (3x+4)(5x+2-(2x-5))
B(x) = (3x+4)(5x+2-2x+5)
B(x) = (3x+4)(3x+7)
5) Résoudre B(x) = 0
B(x) = 0
or B(x) = (3x+4)(3x+7) (question 2)
donc
(3x+4)(3x+7) = 0
2 solutions
3x+4 = 0
3x = -4
x = -4/3
ou
3x+7 = 0
3x = -7
x = -7/3
B(x) admet 2 solutions : x = -4/3 et x = -7/3
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B(x) = (3x+4)(5x+2)-(3x+4)(2x-5)
on met 3x+4 en facteur
B(x) = (3x+4)(5x+2-(2x-5))
B(x) = (3x+4)(5x+2-2x+5)
B(x) = (3x+4)(3x+7)
5) Résoudre B(x) = 0
B(x) = 0
or B(x) = (3x+4)(3x+7) (question 2)
donc
(3x+4)(3x+7) = 0
2 solutions
3x+4 = 0
3x = -4
x = -4/3
ou
3x+7 = 0
3x = -7
x = -7/3
B(x) admet 2 solutions : x = -4/3 et x = -7/3