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LucasBonnet
@LucasBonnet
April 2019
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J'ai un problem avec mon exo de maths.. j'arrive pas a le faire. Vous pouvez m'aider svp??
(3x-2)²=36
(x+V3)²=-4
x³-5x=0
3x(x-1)=5(x-1)
(x+1)²-16x²=0
Merciii
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linouchette
Bonjour
(3x-2)^2 = 36
(3x-2)^2 = 6^2
(3x-2)^2 - 6^2 = 0 c est sous la forme
a^2 - b^2 = (a-b)(a+b)
(3x-2-6)(3x-2+6) = 0
(3x-8)(3x+4) = 0
=> 3x-8 = 0 ou 3x+4 = 0
=> 3x=8 ou 3x=-4
=> x=8/3 ou x=-4/3
(x+v3) ^2 = -4 impossible car un carre
n est jamais négatif
x^3 - 5x = 0
x(x^2-5) = 0 si x=0 ou x=V5
3x (x-1)=5 (x-1)
3x (x-1)-5 (x-1)=0
(x-1)(3x-5)=0 si x=1oux=5/3
(x+1)^2-16^2=0
(x+1-16)(x+1+16)=0 si
x=15 ou x=-17
1 votes
Thanks 1
LucasBonnet
Merrciiii!
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Trouverez-vous la solution de cette suite ?5 + 3 = 289 + 1 = 8108 + 6 = 2145 + 4 = 197 + 3 = ???J'attends des explications ;)
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Lista de comentários
(3x-2)^2 = 36
(3x-2)^2 = 6^2
(3x-2)^2 - 6^2 = 0 c est sous la forme
a^2 - b^2 = (a-b)(a+b)
(3x-2-6)(3x-2+6) = 0
(3x-8)(3x+4) = 0
=> 3x-8 = 0 ou 3x+4 = 0
=> 3x=8 ou 3x=-4
=> x=8/3 ou x=-4/3
(x+v3) ^2 = -4 impossible car un carre
n est jamais négatif
x^3 - 5x = 0
x(x^2-5) = 0 si x=0 ou x=V5
3x (x-1)=5 (x-1)
3x (x-1)-5 (x-1)=0
(x-1)(3x-5)=0 si x=1oux=5/3
(x+1)^2-16^2=0
(x+1-16)(x+1+16)=0 si
x=15 ou x=-17