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Zelomari98
@Zelomari98
May 2019
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Bonsoir les amis je veux votre aide dans cette exercice de mathématiques
résoudre les équations suivante
9x^2 + 6x = -1
( 5x- 4 )^2 - ( 2x- 5 )^2 = 0
( 2x + 1 )^2 = 3
et merci
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loulakar
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Bonsoir,
résoudre les équations suivantes
9x^2 + 6x = -1
9x^2 + 6x + 1 = 0
(3x + 1)^2 = 0
3x + 1 = 0
3x = -1
x = -1/3
( 5x- 4 )^2 - ( 2x- 5 )^2 = 0
(5x - 4 - 2x + 5)(5x - 4 + 2x - 5) = 0
(3x + 1)(7x - 9) = 0
3x + 1 = 0 ou 7x - 9 = 0
3x = -1 ou 7x = 9
x = -1/3 ou x = 9/7
( 2x + 1 )^2 = 3
(2x + 1)^2 - 3 = 0
(2x + 1 - V3)(2x + 1 + V3) = 0
2x + 1 - V3 = 0 ou 2x + 1 + V3 = 0
2x = -1 + V3 ou 2x = -1 - V3
x = (-1 + V3)/2 ou x = (-1 - V3)/2
1 votes
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Killua22
Verified answer
Bonsoir Cher ami voilà la réponse à votre demande
résoudre les équations suivante
9x^2 + 6x = -1
( 3x )² + 2 × 3x × 1 + 1² = 0
( 3x + 1 )² = 0
3x + 1 = 0
3x = -1
x = -1/3
( 5x- 4 )^2 - ( 2x- 5 )^2 = 0
[ ( 5x- 4 ) - ( 2x- 5 ) ] [ ( 5x- 4 ) - ( 2x- 5 ) ] = 0
[ 5x - 4 - 2x + 5 ] [ 5x - 4 + 2x - 5 ] = 0
( 3x + 1 ) ( 7x - 9 ) = 0
3x + 1 = 0. ou 7x - 9 = 0
x = -1/3. ou x = 9/7
( 2x + 1 )^2 = 3
( 2x + 1 )^2 - 3 = 0
( 2x + 1 )^2 - (V3)^2 = 0
[ ( 2x + 1 ) - (V3) ] [ ( 2x + 1 ) + (V3) ] = 0
[ 2x + 1 - V3 ] [ 2x + 1 + V3 ] = 0
2x + 1 - V3 = 0 ou 2x + 1 + V3 = 0
x = ( V3 - 1)/2. ou x = ( - V3 - 1)/2
et j'espère t'avoir aider
0 votes
Thanks 1
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Lista de comentários
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Bonsoir,résoudre les équations suivantes
9x^2 + 6x = -1
9x^2 + 6x + 1 = 0
(3x + 1)^2 = 0
3x + 1 = 0
3x = -1
x = -1/3
( 5x- 4 )^2 - ( 2x- 5 )^2 = 0
(5x - 4 - 2x + 5)(5x - 4 + 2x - 5) = 0
(3x + 1)(7x - 9) = 0
3x + 1 = 0 ou 7x - 9 = 0
3x = -1 ou 7x = 9
x = -1/3 ou x = 9/7
( 2x + 1 )^2 = 3
(2x + 1)^2 - 3 = 0
(2x + 1 - V3)(2x + 1 + V3) = 0
2x + 1 - V3 = 0 ou 2x + 1 + V3 = 0
2x = -1 + V3 ou 2x = -1 - V3
x = (-1 + V3)/2 ou x = (-1 - V3)/2
Verified answer
Bonsoir Cher ami voilà la réponse à votre demanderésoudre les équations suivante
9x^2 + 6x = -1
( 3x )² + 2 × 3x × 1 + 1² = 0
( 3x + 1 )² = 0
3x + 1 = 0
3x = -1
x = -1/3
( 5x- 4 )^2 - ( 2x- 5 )^2 = 0
[ ( 5x- 4 ) - ( 2x- 5 ) ] [ ( 5x- 4 ) - ( 2x- 5 ) ] = 0
[ 5x - 4 - 2x + 5 ] [ 5x - 4 + 2x - 5 ] = 0
( 3x + 1 ) ( 7x - 9 ) = 0
3x + 1 = 0. ou 7x - 9 = 0
x = -1/3. ou x = 9/7
( 2x + 1 )^2 = 3
( 2x + 1 )^2 - 3 = 0
( 2x + 1 )^2 - (V3)^2 = 0
[ ( 2x + 1 ) - (V3) ] [ ( 2x + 1 ) + (V3) ] = 0
[ 2x + 1 - V3 ] [ 2x + 1 + V3 ] = 0
2x + 1 - V3 = 0 ou 2x + 1 + V3 = 0
x = ( V3 - 1)/2. ou x = ( - V3 - 1)/2
et j'espère t'avoir aider