Articles
Register
Sign In
Search
kelya2
@kelya2
January 2021
1
18
Report
Pour demain , l'exercice numéro 2 un prombleme de math svp .. Merci d'avance
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
AhYan
Verified answer
AB = 2x+1
AF = x+3
1) ABCD est un carré de côté 2x+1.
Or aire carré = côté²
Donc A(abcd) = (2x+1)²
ABEF est un rectangle. Or aire d'un rectangle = L×l
Donc A(abef) = (2x+1)(x+3)
2) A(fecd) = A(abcd) - A(abef)
A(fecd) = (2x+1)² - (2x+1)(x+3) = (2x+1)(2x+1-x-3)
A(fecd) = (2x+1)(x-2)
3) l = 2x+1
L = (2x+1)-(x+3) = 2x+1-x-3 = x-2
A(fecd) = (2x+1)×[(2x+1)-(x+3)]
4) A(fecd) = (2x+1)×[(2x+1)-(x+3)] = (2x+1)(2x+1-x-3) = (2x+1)(x-2)
5) On veut :
A(abcd) = 2×A(abef)
(2x+1)² = 2×(2x+1)(x+3)
(2x+1)² - 2×(2x+1)(x+3) = 0
(2x+1)[(2x+1)-2(x+3)] = 0
(2x+1)(2x+1-2x-6) = 0
(2x+1)×(-5) =0
-10x-5 = 0
-10x = 5
x = 5/-10 = -1/2
0 votes
Thanks 0
More Questions From This User
See All
kelya2
January 2021 | 0 Respostas
Responda
kelya2
January 2021 | 0 Respostas
Responda
×
Report "Pour demain , l'exercice numéro 2 un prombleme de math svp .. Merci d'avance.... Pergunta de ideia de kelya2"
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
AB = 2x+1AF = x+3
1) ABCD est un carré de côté 2x+1.
Or aire carré = côté²
Donc A(abcd) = (2x+1)²
ABEF est un rectangle. Or aire d'un rectangle = L×l
Donc A(abef) = (2x+1)(x+3)
2) A(fecd) = A(abcd) - A(abef)
A(fecd) = (2x+1)² - (2x+1)(x+3) = (2x+1)(2x+1-x-3)
A(fecd) = (2x+1)(x-2)
3) l = 2x+1
L = (2x+1)-(x+3) = 2x+1-x-3 = x-2
A(fecd) = (2x+1)×[(2x+1)-(x+3)]
4) A(fecd) = (2x+1)×[(2x+1)-(x+3)] = (2x+1)(2x+1-x-3) = (2x+1)(x-2)
5) On veut :
A(abcd) = 2×A(abef)
(2x+1)² = 2×(2x+1)(x+3)
(2x+1)² - 2×(2x+1)(x+3) = 0
(2x+1)[(2x+1)-2(x+3)] = 0
(2x+1)(2x+1-2x-6) = 0
(2x+1)×(-5) =0
-10x-5 = 0
-10x = 5
x = 5/-10 = -1/2