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Licorne38
@Licorne38
May 2019
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Résoudre dans R, l’équation (3X+1)au carré -16=0.
Merci (c’est l’expression 3X+1 tout entière qui est au carré).
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scoladan
Verified answer
Bonjour,
(3x + 1)² - 16 = 0
⇔ (3x + 1)² - 4² = 0
⇔ [(3x + 1) - 4][(3x + 1) + 4] = 0
⇔ (3x - 3)(3x + 5) = 0
⇒ 3x - 3 = 0 ou 3x + 5 = 0
⇒ x = 1 ou x = -5/3
1 votes
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1D2010
c'est -5/3
1D2010
Verified answer
Bonsoir,
(3x+1)^2-16=0
(3x+1-4)(3x+1+4)=0
(3x-3)(3x+5)=0
3x-3=0 ou 3x+5=0
x=1 ou x=-5/3
Bon courage !
1 votes
Thanks 0
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Lista de comentários
Verified answer
Bonjour,(3x + 1)² - 16 = 0
⇔ (3x + 1)² - 4² = 0
⇔ [(3x + 1) - 4][(3x + 1) + 4] = 0
⇔ (3x - 3)(3x + 5) = 0
⇒ 3x - 3 = 0 ou 3x + 5 = 0
⇒ x = 1 ou x = -5/3
Verified answer
Bonsoir,(3x+1)^2-16=0
(3x+1-4)(3x+1+4)=0
(3x-3)(3x+5)=0
3x-3=0 ou 3x+5=0
x=1 ou x=-5/3
Bon courage !