Home
О нас
Products
Services
Регистрация
Войти
Поиск
Anny1117
@Anny1117
March 2022
2
20
Report
2,3,7 методом интервалов
СРОЧНО!!!!
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms of service
You must agree before submitting.
Send
Answers & Comments
ItalicOpen
2. x1 = 3/2 x2 = -1/3
x ∈ (-∞; -1/3)∪(3/2;∞)
3. x1 = 4/5 x2= -5/2
x ∈ (-∞; -5/2)∪(4/5;∞)
7. x1 = 0 x2= 1 x3= 2 x4= 3
x ∈ (-∞; 0)∪(1;2)∪(3;∞)
1 votes
Thanks 2
300228229
7. x1 = 0 x2= 1 x3= 2 x4=
= 3 x ∈ (-∞; 0)∪(1;2)∪(3;∞)
2. x1 = 3/2 x2 = -1/3
x ∈ (-∞; -1/3)∪(3/2;∞)
3. x1 = 4/5 x2= -5/2
x ∈ (-∞; -5/2)∪(4/5;∞)
1 votes
Thanks 1
300228229
Тот ответ тоже правельный
×
Report "2,3,7 методом интерваловСРОЧНО!!!!..."
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
О нас
Политика конфиденциальности
Правила и условия
Copyright
Контакты
Helpful Social
Get monthly updates
Submit
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
x ∈ (-∞; -1/3)∪(3/2;∞)
3. x1 = 4/5 x2= -5/2
x ∈ (-∞; -5/2)∪(4/5;∞)
7. x1 = 0 x2= 1 x3= 2 x4= 3
x ∈ (-∞; 0)∪(1;2)∪(3;∞)
7. x1 = 0 x2= 1 x3= 2 x4=
= 3 x ∈ (-∞; 0)∪(1;2)∪(3;∞)
2. x1 = 3/2 x2 = -1/3
x ∈ (-∞; -1/3)∪(3/2;∞)
3. x1 = 4/5 x2= -5/2
x ∈ (-∞; -5/2)∪(4/5;∞)