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Hello0412
@Hello0412
May 2019
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Bonjour ! J’ai besoin d’aide pour une question en mathématiques:
Soit f la fonction définie sur IR par f(x)= (2-3x)^2 - (2x+3)^2
Résoudre dans IR l’inéquation f(x) < 0
Merci à ceux qui m’aiderons !
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Stiaen
Verified answer
Bonjour,
Développons d’abord la fonction:
Cherchons les solutions de l'équation:
Dressons le tableau de signe de la fonction:
Résolution de l'équation à partir du tableau:
1 votes
Thanks 2
loulakar
Verified answer
Bonjour,
f(x) = (2 - 3x)^2 - (2x + 3)^2
f(x) = (2 - 3x - 2x - 3)(2 - 3x + 2x + 3)
f(x) = (-5x - 1)(-x + 5)
(-5x - 1)(-x + 5) = 0
-5x - 1 = 0
5x = -1
x = -1/5
-x + 5 = 0
x = 5
x........|...-inf............(-1/5)...........5.............+inf
f(x)....|.............(+).......||.......(-).....||.....(+)..........
f(x) < 0 pour x € ]-1/5;5[
2 votes
Thanks 1
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Lista de comentários
Verified answer
Bonjour,Développons d’abord la fonction:
Cherchons les solutions de l'équation:
Dressons le tableau de signe de la fonction:
Résolution de l'équation à partir du tableau:
Verified answer
Bonjour,f(x) = (2 - 3x)^2 - (2x + 3)^2
f(x) = (2 - 3x - 2x - 3)(2 - 3x + 2x + 3)
f(x) = (-5x - 1)(-x + 5)
(-5x - 1)(-x + 5) = 0
-5x - 1 = 0
5x = -1
x = -1/5
-x + 5 = 0
x = 5
x........|...-inf............(-1/5)...........5.............+inf
f(x)....|.............(+).......||.......(-).....||.....(+)..........
f(x) < 0 pour x € ]-1/5;5[