Bonjour,
1) Développer puis réduire E :
E = (2x-1)²-(2x-1)(x-3)
↳ Identité remarquable : (a-b)² = a²-2ab+b²
E = 4x²-4x+1-(2x-1)(x-3)
E = 4x²-4x+1-(2x²-6x-x+3)
E = 4x²-4x+1-(2x²-7x+3)
E = 4x²-4x+1-2x²+7x-3
E = 2x²-4x+1+7x-3
E = 2x²+3x+1-3
E = 2x²+3x-2
2a) Calculer E pour x = 0 :
E = (2x-1)²-(2x-1)(x-3)
E = (2×0-1)²-(2×0-1)(0-3)
E = (2×0-1)²-(2×0-1)×(-3)
E = (0-1)²-(2×0-1)×(-3)
E = (0-1)²-(0-1)×(-3)
E = (-1)²-(0-1)×(-3)
E = (-1)²-(-1)×(-3)
E = 1-(-1)×(-3)
E = 1+1×(-3)
E = 1+(-3)
E = 1-3
E = -2
2b) Calculer E pour x = 3/4
E = (2x-1)²-(2x-1)(x-3)
E = (2×3/4-1)²-(2×3/4-1)(3/4-3)
E = (3/2-1)²-(2×(3/4)-1)(3/4-3)
E = (1/2)²-(2×(3/4)-1)(3/4-3)
E = 1/4-(2×(3/4)-1)(3/4-3)
E = 1/4-(3/2-1)(3/4-3)
E = 1/4-1/2(3/4-3)
E = 1/4-1/2(-9/4)
E = 1/4-(-9/8)
E = 1/4+9/8
E = 11/8
Démontrer que cette affirmation et vraie :
= (n+1)²-(n-1)²
= n²+2n+1-(n²-2n+1)
= n²+2n+1-n²+2n-1
Il nous reste 4n
C'est un multiple de 4 donc l'affirmation est approuvé ;)
Bonne journée.