Articles
Register
Sign In
Search
wendy14
@wendy14
January 2021
1
64
Report
bonsoir je voudrai de l'aide pour l'exercice de math . 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
anylor
Bonsoir
3.1)
|x+7|≥3
1er cas
x+7 ≥ 0 => x+7 ≥3 => x ≥ 3-7 => x ≥ - 4
2nd cas
x+7 ≤ 0 => - (x +7 ) ≥ 3 => -x -7≥3 => x ≤ -7-3 => x ≤ -10
solution :
x≤ -10 OU x ≥ -4
3.2)
|3x-1| ≤5/2
1er cas
3x-1≥0 => 3x-1 ≤ 5/2 => 3x ≤ 5/2 +2/2 => 3x ≤ 7/2
x ≤ 7/6
2nd cas
3x-1≤0 => -(3x-1) ≤ 5/2 => -3x +1 ≤ 5/2
-3x ≤ 5/2 -2/2
-3x ≤ 3/2
3x ≥ -3/2
x ≥ - 3/6 => x ≥ -1/2 ( on simplifie 3/6 = 1/2)
solution
-1/2 ≤ x ≤ 7/6
3.3)
|x+6| ≤0
1er cas
x+6 ≤0 => x ≤ -6
2nd cas
-x-6 ≤0 => x ≥-6
1 seule solution possible x= - 6
3. 4)
|x| +x = 0
1er cas
x>0
x+x = 0 => 2x =0 => x =0
2nd cas
si x <0
-x + x = 0 0 = 0 toujours vrai
donc solution :
x ≤ 0
3.5)
il faut faire un tableau avec les différentes valeurs des valeurs absolues
solution: x=6 ou x =2/3
3.6)
1er cas
x²-4x-5≥1
x²-4x-6≥1
utiliser méthode delta pour factoriser
x≤2-√10 ou x≥2+√10
2nd cas
-(x²- 4x-5) ≥1
-x²+ 4x +4≥0
2 - 2√2 ≤ x ≤ 2 + 2√2
en définitive
solution :
2 - 2
√2 ≤ x ≤ 2 + 2√2
2 votes
Thanks 1
More Questions From This User
See All
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Bonsoir j'ai un exercice de math sur les vecteur et je n'arrive pas c'est pour deman c'est urgent merci de votre aide en avance. l'exercice est sur le document ci-joint.
Responda
wendy14
January 2021 | 0 Respostas
Responda
wendy14
January 2021 | 0 Respostas
Responda
×
Report "bonsoir je voudrai de l'aide pour l'exercice de math . merci d'avance.... Pergunta de ideia de wendy14"
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
3.1)
|x+7|≥3
1er cas
x+7 ≥ 0 => x+7 ≥3 => x ≥ 3-7 => x ≥ - 4
2nd cas
x+7 ≤ 0 => - (x +7 ) ≥ 3 => -x -7≥3 => x ≤ -7-3 => x ≤ -10
solution :
x≤ -10 OU x ≥ -4
3.2)
|3x-1| ≤5/2
1er cas
3x-1≥0 => 3x-1 ≤ 5/2 => 3x ≤ 5/2 +2/2 => 3x ≤ 7/2
x ≤ 7/6
2nd cas
3x-1≤0 => -(3x-1) ≤ 5/2 => -3x +1 ≤ 5/2
-3x ≤ 5/2 -2/2
-3x ≤ 3/2
3x ≥ -3/2
x ≥ - 3/6 => x ≥ -1/2 ( on simplifie 3/6 = 1/2)
solution
-1/2 ≤ x ≤ 7/6
3.3)
|x+6| ≤0
1er cas
x+6 ≤0 => x ≤ -6
2nd cas
-x-6 ≤0 => x ≥-6
1 seule solution possible x= - 6
3. 4)
|x| +x = 0
1er cas
x>0
x+x = 0 => 2x =0 => x =0
2nd cas
si x <0
-x + x = 0 0 = 0 toujours vrai
donc solution :
x ≤ 0
3.5)
il faut faire un tableau avec les différentes valeurs des valeurs absolues
solution: x=6 ou x =2/3
3.6)
1er cas
x²-4x-5≥1
x²-4x-6≥1
utiliser méthode delta pour factoriser
x≤2-√10 ou x≥2+√10
2nd cas
-(x²- 4x-5) ≥1
-x²+ 4x +4≥0
2 - 2√2 ≤ x ≤ 2 + 2√2
en définitive
solution :
2 - 2√2 ≤ x ≤ 2 + 2√2