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Unesimplefille
@Unesimplefille
May 2019
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Bonjour vous pourriez m'aider pour un exercice d'un dm de maths s'il vous plait. A l'aide des identités remarquables développer et réduire A=(2n-3)*(2n+3)
En déduire le resultat de 197x203, expliquez le raisonnement.
Merci beaucoup pour votre aide.
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trudelmichel
Verified answer
Bonjour
(2n+3)(2n-3)=4n²+6n-6n-9=4n²-9
(2n+3)(2n-3) correspond a (a+b)(a-b)=a²-b²
(2n+3)(2n-3=4n²-9
197=200-3 203=200+3
197x203=(200+3)(200-3)
(200+3)(200-3) correspond à (a+b)(a-b)=a²-b²
(200+3)(200-3)=(200)²-3²
40000-9=39991
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ficanas06
Verified answer
A=(2n-3)*(2n+3) est du type de l'identité remarquable a²-b²= (a-b)(a+b)
donc
A=(2n-3)*(2n+3) =
4n²-9
197 * 203 = (2*100 - 3) (2*100 + 3) = 4*10 000 - 9 = 40 000 - 9 =
39 991
0 votes
Thanks 0
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Verified answer
Bonjour(2n+3)(2n-3)=4n²+6n-6n-9=4n²-9
(2n+3)(2n-3) correspond a (a+b)(a-b)=a²-b²
(2n+3)(2n-3=4n²-9
197=200-3 203=200+3
197x203=(200+3)(200-3)
(200+3)(200-3) correspond à (a+b)(a-b)=a²-b²
(200+3)(200-3)=(200)²-3²
40000-9=39991
Verified answer
A=(2n-3)*(2n+3) est du type de l'identité remarquable a²-b²= (a-b)(a+b)donc A=(2n-3)*(2n+3) = 4n²-9
197 * 203 = (2*100 - 3) (2*100 + 3) = 4*10 000 - 9 = 40 000 - 9 = 39 991