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Inesabd1904
@Inesabd1904
January 2021
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Bonsoir,
Aidez moi s'il vous plait je ne sais pas comment faire avec ce problème ouvert en maths.
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scoladan
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Bonjour,
on pose AB = x
Donc BC = x + 3
Et, car ABC est rectangle en A :
AB^2 + AC^2 = BC^2
soit AC^2 = BC^2 - AB^2
==> AC^2 = (x+3)^2 - x^2
<=> AC^2 = 6x + 9
On veut AC >ou= x
==> AC^2 >ou= x^2
<=> 6x + 9 >ou= x^2
Soit x^2 - 6x - 9 <ou= 0
Delta = 72 = (6Racine(2))^2
Donc 2 solutions :
x1 = (6-6V(2))/2 négatif --> on en tient pas compte
x2 = (6+6V(2))/2 = 3(1+V2)
Le trinôme sera négatif pour x appartenant à [0, 3(1+V(2))]
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Inesabd1904
Merci infiniment !! :D
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Verified answer
Bonjour,on pose AB = x
Donc BC = x + 3
Et, car ABC est rectangle en A :
AB^2 + AC^2 = BC^2
soit AC^2 = BC^2 - AB^2
==> AC^2 = (x+3)^2 - x^2
<=> AC^2 = 6x + 9
On veut AC >ou= x
==> AC^2 >ou= x^2
<=> 6x + 9 >ou= x^2
Soit x^2 - 6x - 9 <ou= 0
Delta = 72 = (6Racine(2))^2
Donc 2 solutions :
x1 = (6-6V(2))/2 négatif --> on en tient pas compte
x2 = (6+6V(2))/2 = 3(1+V2)
Le trinôme sera négatif pour x appartenant à [0, 3(1+V(2))]