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Juliendemonaco
@Juliendemonaco
April 2019
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HELP PLEASE!!
D= (2x+1)au carré -(x-3)(2x+1)
RÉSOUDRE L ÉQUATION D = 0
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Bonjour,
(2x + 1)² - (x - 3)(2x + 1) = 0
(2x + 1)(2x + 1) - (x - 3)(2x + 1) = 0
(2x + 1) [(2x + 1) - (x - 3)] = 0
(2x + 1) (2x + 1 - x + 3) = 0
(2x + 1) (x + 4) = 0
2x + 1 = 0 ou x + 4 = 0
2x = -1 ou x = -4
x = -1/2 ou x = -4
S = {-1/2 ; -4}
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nonoduchablais
D= (2x+1)² -(x-3)(2x+1)
= 0
D = (2x + 1)(2x +1) - (x - 3)(2x + 1) = 0
D = (2x + 1)[(2x + 1) - (x - 3)] = 0
D = (2x + 1)(2x + 1 - x + 3) = 0
D = (2x + 1)(x + 4) = 0
Comme ce produit est nul alors l'un moins de ses facteurs est nul:
2x + 1 = 0 ou x + 4 = 0
2x = (-1) x = (-4)
x = (-1/2)
Les deux solutions de cette équations sont (-1/2) et (-4)
J'espère t'avoir aidé! :)
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(2x + 1)² - (x - 3)(2x + 1) = 0
(2x + 1)(2x + 1) - (x - 3)(2x + 1) = 0
(2x + 1) [(2x + 1) - (x - 3)] = 0
(2x + 1) (2x + 1 - x + 3) = 0
(2x + 1) (x + 4) = 0
2x + 1 = 0 ou x + 4 = 0
2x = -1 ou x = -4
x = -1/2 ou x = -4
S = {-1/2 ; -4}
D = (2x + 1)(2x +1) - (x - 3)(2x + 1) = 0
D = (2x + 1)[(2x + 1) - (x - 3)] = 0
D = (2x + 1)(2x + 1 - x + 3) = 0
D = (2x + 1)(x + 4) = 0
Comme ce produit est nul alors l'un moins de ses facteurs est nul:
2x + 1 = 0 ou x + 4 = 0
2x = (-1) x = (-4)
x = (-1/2)
Les deux solutions de cette équations sont (-1/2) et (-4)
J'espère t'avoir aidé! :)