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Tartouille
@Tartouille
April 2019
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Factorisation des 4 expressions suivantes SVP - Merci
(5x-8)² + (5x-8)
(2x+5)² - 81
(2x+7)² - 64
4x² - 40x + 100
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kittydouce
Vérifie avant de tout recopier:
(5x-8)² + (5x-8)
(5x-8)x[(5x-8)+1]
(5x-8)x(5x-8+1)
(5x-8)x(5x-7)
(2x+5)² - 81
(2x+5)²-9²
(2x+5+9)(2x+5-9)
(2x+14)(2x-4)
(2x+7)² - 64
(2x+7)²-8²
(2x+7+8)(2x+7-8)
(2x+15)(2x-1)
4x² - 40x + 100
(2x-10)²
0 votes
Thanks 1
tartouille
Kittydouce, c'est toi la meilleure réponse, j'ai mal cliqué !!!!!! oups
lulufabdidith
(5x-8)²+(5x-8)
facteur commun (5x-8)
donc (5x-8)(5x-8)
factoriser
(2x + 5)² -81
(2x + 5)² -81 = 4x² +20x+25-81
4x² + 20x -56 = 4(x² +5x-14)
factoriser
(2x+7)²-64
(2x+7)²-64=4x²+28x+49-64
4x²+28x-15= 4(x²+7x-3.75)
1 votes
Thanks 1
tartouille
Merci Kittydouce je vais refaire pour bien comprendre
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Lista de comentários
(5x-8)² + (5x-8)
(5x-8)x[(5x-8)+1]
(5x-8)x(5x-8+1)
(5x-8)x(5x-7)
(2x+5)² - 81
(2x+5)²-9²
(2x+5+9)(2x+5-9)
(2x+14)(2x-4)
(2x+7)² - 64
(2x+7)²-8²
(2x+7+8)(2x+7-8)
(2x+15)(2x-1)
4x² - 40x + 100
(2x-10)²
facteur commun (5x-8)
donc (5x-8)(5x-8)
factoriser
(2x + 5)² -81
(2x + 5)² -81 = 4x² +20x+25-81
4x² + 20x -56 = 4(x² +5x-14)
factoriser
(2x+7)²-64
(2x+7)²-64=4x²+28x+49-64
4x²+28x-15= 4(x²+7x-3.75)