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laura81
@laura81
June 2021
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Bonjour, pouvez vous m'aidez svp ? Merci
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isapaul
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Bonsoir
f(x) = (x-1)(x+3) et g(x) = (2-x)(x-1)
f(x) - g(x) = (x-1)(2x+1)
f(x) = g(x) revient à
f(x) - g(x) = 0 donc
(x-1)(2x+1) = 0 produit de facteurs est nul si un fzvteur est nul
soit x - 1 = 0 pour x= 1
soit 2x + 1 = 0 pour x = -1/2
f(x) = -4x - 12 alors
(x-1)(x+3) = -4x - 12
x² + 3x - x - 3 = -4x - 12
x² + 2x - 3 = -4x - 12
x² +6x + 9 = 0
(x+3)²= 0
x = -3
Bonne soirée
2 votes
Thanks 1
laura81
Comment on fait dans la question d pour contrôler graphiquement le résultat ?
laura81
Et comment on fait pour la b svp ? Je n'ai pas compris
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Bonsoirf(x) = (x-1)(x+3) et g(x) = (2-x)(x-1)
f(x) - g(x) = (x-1)(2x+1)
f(x) = g(x) revient à
f(x) - g(x) = 0 donc
(x-1)(2x+1) = 0 produit de facteurs est nul si un fzvteur est nul
soit x - 1 = 0 pour x= 1
soit 2x + 1 = 0 pour x = -1/2
f(x) = -4x - 12 alors
(x-1)(x+3) = -4x - 12
x² + 3x - x - 3 = -4x - 12
x² + 2x - 3 = -4x - 12
x² +6x + 9 = 0
(x+3)²= 0
x = -3
Bonne soirée