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Melissahfrch
@Melissahfrch
May 2019
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Bonjour pouvez vous m'aider à résoudre cette équation s'il vous plaît :
(2y+3)²=(4y-1)(y+5)
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maudmarine
Verified answer
Bonjour,
(2y+3)²=(4y-1)(y+5)
4y² + 12y + 9 = 4y² + 20y - y - 5
4y² - 4y² + 12y - 20y + y = - 5 - 9
- 7y = - 14
y = 14/7
y = 2
1 votes
Thanks 1
nini999
Verified answer
Bonjour :
(2y + 3)² = (4y - 1)(y + 5)
(2y + 3)² il prend la forme de (a+b)² = a²+2ab+b²
Alors :
4y² + 12y + 9 = 4y² + 20y - y - 5
4y² + 12y + 9 - 4y² - 20y + y + 5 = 0
4y² - 4y² + 12y - 20y + y + 9 + 5 = 0
0 + -7y + 14 = 0
-7y + 14 = 0
-7y = -14
y = -14/-7
y = 14/7
y = 2
J’espère t'avoir t'aider
1 votes
Thanks 1
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Lista de comentários
Verified answer
Bonjour,(2y+3)²=(4y-1)(y+5)
4y² + 12y + 9 = 4y² + 20y - y - 5
4y² - 4y² + 12y - 20y + y = - 5 - 9
- 7y = - 14
y = 14/7
y = 2
Verified answer
Bonjour :(2y + 3)² = (4y - 1)(y + 5)
(2y + 3)² il prend la forme de (a+b)² = a²+2ab+b²
Alors :
4y² + 12y + 9 = 4y² + 20y - y - 5
4y² + 12y + 9 - 4y² - 20y + y + 5 = 0
4y² - 4y² + 12y - 20y + y + 9 + 5 = 0
0 + -7y + 14 = 0
-7y + 14 = 0
-7y = -14
y = -14/-7
y = 14/7
y = 2
J’espère t'avoir t'aider