Bonjour, pouvez-vous vérifiez mes résultats SVP ? Merci d'avance ! Factorisez les expressions suivantes : A= (2t+1)²-(t+1)² = [(2t+1)(t+1)] [(2t+1)(t-1)] = (3t+
Mathématiques
Salut437
Question
Bonjour, pouvez-vous vérifiez mes résultats SVP ? Merci d'avance !
Factorisez les expressions suivantes :
A= (2t+1)²-(t+1)²
= [(2t+1)(t+1)] [(2t+1)(t-1)]
= (3t+2) 3t
B= (3t-7)²-25
= [(3t-7)+25] [(3t-7)-25]
= (3t-7+25) (3t-7-25)
= (3t+18) (3t-32)
C= (3t+4)²-(2t+3)²
=[(3t+4)(2t+3] [(3t+4)(2t-3)]
= (5t+7) (5t+1)
Factorisez les expressions suivantes :
A= (2t+1)²-(t+1)²
= [(2t+1)(t+1)] [(2t+1)(t-1)]
= (3t+2) 3t
B= (3t-7)²-25
= [(3t-7)+25] [(3t-7)-25]
= (3t-7+25) (3t-7-25)
= (3t+18) (3t-32)
C= (3t+4)²-(2t+3)²
=[(3t+4)(2t+3] [(3t+4)(2t-3)]
= (5t+7) (5t+1)
1 Réponse
-
1. Réponse danielwenin
A= (2t+1)²-(t+1)²
= [(2t+1) + (t+1)] [(2t+1) - (t-1)]
= (3t+2)( t + 2 )
B= (3t-7)²-25
= (3t - 7 - 5)(3t - 7 +5)
= (3t - 12)(3t-2) = 3(t-4)(3t-2)
C= (3t+4)²-(2t+3)²
=[(3t+4)-(2t+3)] [(3t+4)+(2t-3)]
= (t+1)(5t+1)
tu as toujours commis le même erreur
a² - b² = (a - b)(a + b)