Articles
Register
Sign In
Search
Laura97
@Laura97
April 2019
2
52
Report
Bonjour, qui pourrez nous aidé en mathématiques niveau 3e ? C'est pour mon frère. Merci d'avance.
Développer et réduire les expressions suivantes :
A= -2(7x+4)²
B= (3x-2)(3x+2)+(2x-5)²
C= 3(x+3)²-(2x-1(3x+1)
Exercice 2 :
Factoriser les expressions suivantes
E = 9x²-24x+16
F = (x+1) (x+3)-5(x+3)
G = 100t²-121
H = (4x+1)(7x+2)+(4x+1)²
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
winner123
Verified answer
Bonjour
A = - 2 ( 49 x² + 56 x + 16)
A = - 98 x² - 112 x - 32
B = 9 x² + 6 x - 6 x - 4 + 4 x² - 20 x + 25
B = 13 x² - 20 x + 21
c = 3( x² + 6 x +9) - ( 6 x² + 2 x - 3x - 1)
C = 3 x² + 18 x + 27 - 6 x² - 2 x + 3 x + 1
C = - 3 x² + 19 x + 28
E = ( 3 x - 4)²
F = (x +3) ( x +1 - 5)
f = ( x +3) ( x -4)
G = ( 10 t - 11) ( 10 t +11)
h = ( 4 x +1) ( 7 x +2 + 4 x +1)
H = ( 4 x +1) ( 11 x + 5)
1 votes
Thanks 1
nathalienouts
Développer et réduire les expressions suivantes :
A= -2(7x+4)²
A = -2(49x²+56x+16)
A = -98x²-112x-32
B= (3x-2)(3x+2)+(2x-5)²
B= (9x²+6x-6x-4)+(4x²-20x+25)
B = 9x²-4+4x²-20x+25
B = 13x²-20x+21
C= 3(x+3)²-(2x-1)(3x+1)
C = 3(x²+6x+9)-(6x²+2x-3x-1)
C = 3x²+18x+27-6x²+x+1
C = -3x²+19x+28
Exercice 2 :
Factoriser les expressions suivantes
E = 9x²-24x+16 identité remarquable (a+b)² = a²+2ab+b²
E = (3x-4)²
F = (x+1) (x+3)-5(x+3) on met x+3 en facteur
F = (x+3)(x+1-5)
F = (x+3)(x-4)
G = 100t²-121 identité remarquable a²-b² = (a+b)(a-b)
G =(10t)²-11²
G = (10t+11)(10t-11)
H = (4x+1)(7x+2)+(4x+1)² on met 4x+1 en facteur
H = (4x+1)(7x+2+4x+1)
H = (4x+1)(11x+3)
2 votes
Thanks 1
More Questions From This User
See All
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
Laura97
June 2021 | 0 Respostas
Responda
×
Report "Bonjour, qui pourrez nous aidé en mathématiques niveau 3e ? C'est pour mon frère. Merci d'avance. Dé.... Pergunta de ideia de Laura97"
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
BonjourA = - 2 ( 49 x² + 56 x + 16)
A = - 98 x² - 112 x - 32
B = 9 x² + 6 x - 6 x - 4 + 4 x² - 20 x + 25
B = 13 x² - 20 x + 21
c = 3( x² + 6 x +9) - ( 6 x² + 2 x - 3x - 1)
C = 3 x² + 18 x + 27 - 6 x² - 2 x + 3 x + 1
C = - 3 x² + 19 x + 28
E = ( 3 x - 4)²
F = (x +3) ( x +1 - 5)
f = ( x +3) ( x -4)
G = ( 10 t - 11) ( 10 t +11)
h = ( 4 x +1) ( 7 x +2 + 4 x +1)
H = ( 4 x +1) ( 11 x + 5)
A= -2(7x+4)²
A = -2(49x²+56x+16)
A = -98x²-112x-32
B= (3x-2)(3x+2)+(2x-5)²
B= (9x²+6x-6x-4)+(4x²-20x+25)
B = 9x²-4+4x²-20x+25
B = 13x²-20x+21
C= 3(x+3)²-(2x-1)(3x+1)
C = 3(x²+6x+9)-(6x²+2x-3x-1)
C = 3x²+18x+27-6x²+x+1
C = -3x²+19x+28
Exercice 2 :
Factoriser les expressions suivantes
E = 9x²-24x+16 identité remarquable (a+b)² = a²+2ab+b²
E = (3x-4)²
F = (x+1) (x+3)-5(x+3) on met x+3 en facteur
F = (x+3)(x+1-5)
F = (x+3)(x-4)
G = 100t²-121 identité remarquable a²-b² = (a+b)(a-b)
G =(10t)²-11²
G = (10t+11)(10t-11)
H = (4x+1)(7x+2)+(4x+1)² on met 4x+1 en facteur
H = (4x+1)(7x+2+4x+1)
H = (4x+1)(11x+3)