Articles
Register
Sign In
Search
alicia56bzh
@alicia56bzh
January 2021
1
27
Report
bonjours besoin d'aide pour la partie 2 de cet exo svp
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
nguyenso9
Verified answer
Bonjour,
8) f(-1) = (-1)² - 4 = 1 - 4 = 3
f(0) = 0² - 4 = 0 - 4 = -4
f(√2) = (√2)² - 4 = 2 - 4 = -2
9) x² - 4 = 0
Par lecture graphique, les antecedents de 0 peuvent etre -2 et 2
x² - 4 = 5
x² - 4 - 5 = 0
x² - 9 = 0
a = 1, b = 0, c = -9
Δ = 0² - 4*1*(-9) = 0 - (-36) = 0 + 36 = 36
Δ > 0, donc l'equation x² - 9 = 0 admet 2 solutions
x1 = (0 - √36)/2 = -6/2 = -3
x2 = (0 + √36)/2 = 6/2 = 3
Les antecedents de 5 peuvent etre -3 et 3
x² - 4 = -5
x² - 4 + 5 = 0
x² + 1 = 0
a = 1, b = 0, c = 1
Δ = b² - 4ac = 0² - 4*1*1 = 0 - 4 = -4
Δ < 0 donc l'equation x² + 1 = 0 n'admet pas de solution.
Il n'y a pas d'antecedent de -5.
10) f(x) = 1
x² - 4 = 1
x² - 4 - 1 = 0
x² - 5 = 0
a = 1, b = 0, c = -5
Δ = b² - 4ac = 0² - 4*1*(-5) = 0 + 20 = 20
Δ > 0 donc l'equation x² - 5 = 0 admet deux solutions
x1 = (0 - √20)/2 = -√20/2 = -√5
x² = (0 + √20)/2 = √20/2 = √5
Les solutions de f(x) = 1 sont -√5 et √5.
0 votes
Thanks 0
alicia56bzh
merci bien !
More Questions From This User
See All
alicia56bzh
January 2021 | 0 Respostas
Responda
alicia56bzh
January 2021 | 0 Respostas
bonjours je suis en seconde, jai besoin d'aide pour la question 5 de la partie 1 svp
Responda
Alicia56bzh
May 2019 | 0 Respostas
Bonjour j'ai vraiment besoin d'aide svp je suis en 3eme et j'ai un devoir d'histoire et je n'arrive pas la question 2a et b
Responda
Alicia56bzh
May 2019 | 0 Respostas
Responda
×
Report "bonjours besoin d'aide pour la partie 2 de cet exo svp.... Pergunta de ideia de alicia56bzh"
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,8) f(-1) = (-1)² - 4 = 1 - 4 = 3
f(0) = 0² - 4 = 0 - 4 = -4
f(√2) = (√2)² - 4 = 2 - 4 = -2
9) x² - 4 = 0
Par lecture graphique, les antecedents de 0 peuvent etre -2 et 2
x² - 4 = 5
x² - 4 - 5 = 0
x² - 9 = 0
a = 1, b = 0, c = -9
Δ = 0² - 4*1*(-9) = 0 - (-36) = 0 + 36 = 36
Δ > 0, donc l'equation x² - 9 = 0 admet 2 solutions
x1 = (0 - √36)/2 = -6/2 = -3
x2 = (0 + √36)/2 = 6/2 = 3
Les antecedents de 5 peuvent etre -3 et 3
x² - 4 = -5
x² - 4 + 5 = 0
x² + 1 = 0
a = 1, b = 0, c = 1
Δ = b² - 4ac = 0² - 4*1*1 = 0 - 4 = -4
Δ < 0 donc l'equation x² + 1 = 0 n'admet pas de solution.
Il n'y a pas d'antecedent de -5.
10) f(x) = 1
x² - 4 = 1
x² - 4 - 1 = 0
x² - 5 = 0
a = 1, b = 0, c = -5
Δ = b² - 4ac = 0² - 4*1*(-5) = 0 + 20 = 20
Δ > 0 donc l'equation x² - 5 = 0 admet deux solutions
x1 = (0 - √20)/2 = -√20/2 = -√5
x² = (0 + √20)/2 = √20/2 = √5
Les solutions de f(x) = 1 sont -√5 et √5.