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Siham200
@Siham200
May 2019
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Aidez moi s'il vous plait!!
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aymanemaysae
Bonsoir ;
Exercice n° 1 :
Posons P(x) = ax² + bx + c avec a , b et c des nombres réels .
P(x+1) = a(x+1)² + b(x+1) + c
= a(x² + 2x + 1) + bx + b + c
= ax² + 2ax + a + bx + b + c
= ax² + (2a + b)x + a + b + c
donc : P(x+1) - P(x) = ax² + (2a + b)x + a + b + c - ax² - bx - c
= 2ax + a + b = x ,
donc : 2a = 1 et a + b = 0
donc : a = 1/2 et b = - a = - 1/2 ,
donc : P(x) = 1/2 x² - 1/2 x + c avec c ∈ R .
1 + 2 + ............ + n = n(n+1)/2 ,
donc : 1 + 2 + ............. + 2005 = 2005 x (2005 + 1)/2 = 2005 x 2006/2
= 2005 x 1003 = 2011015 .
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Siham200
merci beaucoup mais j'ai pas bien comprie le deuxiéme exercice
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Exercice n° 1 :
Posons P(x) = ax² + bx + c avec a , b et c des nombres réels .
P(x+1) = a(x+1)² + b(x+1) + c
= a(x² + 2x + 1) + bx + b + c
= ax² + 2ax + a + bx + b + c
= ax² + (2a + b)x + a + b + c
donc : P(x+1) - P(x) = ax² + (2a + b)x + a + b + c - ax² - bx - c
= 2ax + a + b = x ,
donc : 2a = 1 et a + b = 0
donc : a = 1/2 et b = - a = - 1/2 ,
donc : P(x) = 1/2 x² - 1/2 x + c avec c ∈ R .
1 + 2 + ............ + n = n(n+1)/2 ,
donc : 1 + 2 + ............. + 2005 = 2005 x (2005 + 1)/2 = 2005 x 2006/2
= 2005 x 1003 = 2011015 .