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Anisovich
@Anisovich
May 2019
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Bonjour pouvez vous m'aidez c pour lundi merci d'avance
développer réduire
a=(5x-1)(2x-3)
b=(5x-1)²+(2x+3)
c=(2x-1)(x+6)-(4x+3)(4x-3)
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PAU64
Verified answer
A = (5x - 1) (2x - 3)
A = 5x * 2x + 5x * (- 3) - 1 * 2x - 1 * (- 3)
A = 10x² - 15x - 2x + 3
A = 10x² - 17x + 3
B = (5x - 1)² + (2x + 3)
B = (5x)² - 2 * 5x * 1 + 1² + 2x + 3
B = 25x² - 10x + 1 + 2x + 3
B = 25x² - 8x + 4
C = (2x - 1) (x + 6) - (4x + 3) (4x - 3)
C = [2x * x + 2x * 6 - 1 * x - 1 * 6] - [(4x)² - 3²]
C = 2x² + 12x - x - 6 - (16x² - 9)
C = 2x² + 12x - x - 6 - 16x² + 9
C = - 14x² + 11x + 3
3 votes
Thanks 1
anisovich
oui exactement
PAU64
E = (5t + 6)²
anisovich
nn c sa
PAU64
D = (2x - 5) [(2x - 5) + 3] ; donc D = (2x - 5) (2x - 2)
anisovich
a ok j'avais pas trouver sa mai bon je vai relir pour comprendre
anisovich
ha ok la réponse de E c'est (5t+6)²
PAU64
Oui c'est ça :)
anisovich
merci ;)
anisovich
tu peu me détaillez la E
anisovich
SVP :)
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Verified answer
A = (5x - 1) (2x - 3)A = 5x * 2x + 5x * (- 3) - 1 * 2x - 1 * (- 3)
A = 10x² - 15x - 2x + 3
A = 10x² - 17x + 3
B = (5x - 1)² + (2x + 3)
B = (5x)² - 2 * 5x * 1 + 1² + 2x + 3
B = 25x² - 10x + 1 + 2x + 3
B = 25x² - 8x + 4
C = (2x - 1) (x + 6) - (4x + 3) (4x - 3)
C = [2x * x + 2x * 6 - 1 * x - 1 * 6] - [(4x)² - 3²]
C = 2x² + 12x - x - 6 - (16x² - 9)
C = 2x² + 12x - x - 6 - 16x² + 9
C = - 14x² + 11x + 3