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enihs
@enihs
January 2021
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Bonjour j'aimerais de l'aide pour cette question : Factorisez 3n^2+7n+4 puis en deduire le PGCD de ((n+1)^2;3n^2+7n+4)
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aymanemaysae
Bonsoir ;
On a : 3n² + 7n + 4 = 3n² + 3n + 4n + 4 = 3n(n + 1) + 4(n + 1)
= (n + 1)(3n + 4) .
On a aussi : PGCD((n + 1)² ; 3n² + 7n + 4) = PGCD((n + 1)² ; (n + 1)(3n + 4))
= (n + 1) PGCD(n + 1 ; 3n + 4) = (n + 1) PGCD(n + 1 ; 3(n + 1) + 1)
= n + 1 car PGCD(n + 1 ; 3(n + 1) + 1) = 1 .
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Thanks 1
enihs
D'accord merci... Il fallait vraiment le trouver...
aymanemaysae
De rien .
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On a : 3n² + 7n + 4 = 3n² + 3n + 4n + 4 = 3n(n + 1) + 4(n + 1)
= (n + 1)(3n + 4) .
On a aussi : PGCD((n + 1)² ; 3n² + 7n + 4) = PGCD((n + 1)² ; (n + 1)(3n + 4))
= (n + 1) PGCD(n + 1 ; 3n + 4) = (n + 1) PGCD(n + 1 ; 3(n + 1) + 1)
= n + 1 car PGCD(n + 1 ; 3(n + 1) + 1) = 1 .