Articles
Register
Sign In
Search
Inesabd1904
@Inesabd1904
January 2021
1
16
Report
Bonsoir,
Aidez moi s'il vous plait je ne sais pas comment faire avec ce problème ouvert en maths.
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
scoladan
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))]
0 votes
Thanks 0
Inesabd1904
Merci infiniment !! :D
More Questions From This User
See All
Inesabd1904
January 2021 | 0 Respostas
Responda
Inesabd1904
January 2021 | 0 Respostas
Responda
Inesabd1904
January 2021 | 0 Respostas
Responda
Inesabd1904
January 2021 | 0 Respostas
Responda
Inesabd1904
May 2019 | 0 Respostas
Responda
×
Report "Bonsoir,Aidez moi s'il vous plait je ne sais pas comment faire avec ce problème ouvert en maths.... Pergunta de ideia de Inesabd1904"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
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))]