Home
О нас
Products
Services
Регистрация
Войти
Поиск
Grafkom
@Grafkom
July 2021
1
8
Report
найти сумму наибольшего целого и наименьшего целого решений неравенств:
[tex] \frac{|x-30|(x^2+9)}{|x-7|-10} \leq 0[/tex]
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
sedinalana
Verified answer
X²+9>0 при любом х⇒|x-30|/(|x-7|-10)≤0
1)x<7
(30-x)/(7-x-10)≤0
(x-30)/(x+3)≤0
x=30 x=-3
-3<x<30
x∈(-3;7)
2)7≤x<30
(30-x)/(x-7-10)≤0
(x-30)/(x-17)≥0
x=30 x=17
x<17 U x>30
x∈[7;17)
3)x≥30
(x-30)/(x-17)≤0
x=30 x=17
17<x≤30
x=30
Ответ x∈(-3;17) U {30}
0 votes
Thanks 0
рекомендуемые вопросы
rarrrrrrrr
August 2022 | 0 Ответы
o chem dolzhny pozabotitsya v pervuyu ochered vzroslye pri organizacionnom vyvoze n
danilarsentev
August 2022 | 0 Ответы
est dva stanka na kotoryh vypuskayut odinakovye zapchasti odin proizvodit a zapcha
myachina8
August 2022 | 0 Ответы
najti po grafiku otnoshenie v3v1 v otvetah napisano 9 no nuzhno reshenie
ydpmn7cn6w
August 2022 | 0 Ответы
Choose the correct preposition: 1.I am fond (out,of,from) literature. 2.where ar...
millermilena658
August 2022 | 0 Ответы
opredelite kak sozdavalas i kto sozdaval arabskoe gosudarstvo v kracii
MrZooM222
August 2022 | 0 Ответы
ch ajtmanov v rasskaze krasnoe yabloko ispolzuet metod rasskaz v rasskaze opi
timobila47
August 2022 | 0 Ответы
kakovo bylo naznachenie kazhdoj iz chastej vizantijskogo hrama pomogite pozhalujsta
ivanyyaremkiv
August 2022 | 0 Ответы
moment. 6....
pozhalujsta8b98a56c0152a07b8f4cbcd89aa2f01e 97513
sarvinozwakirjanova
August 2022 | 0 Ответы
pomogite pozhalusto pzha519d7eb8246a08ab0df06cc59e9dedb 6631
×
Report "tex]..."
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 © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
X²+9>0 при любом х⇒|x-30|/(|x-7|-10)≤01)x<7
(30-x)/(7-x-10)≤0
(x-30)/(x+3)≤0
x=30 x=-3
-3<x<30
x∈(-3;7)
2)7≤x<30
(30-x)/(x-7-10)≤0
(x-30)/(x-17)≥0
x=30 x=17
x<17 U x>30
x∈[7;17)
3)x≥30
(x-30)/(x-17)≤0
x=30 x=17
17<x≤30
x=30
Ответ x∈(-3;17) U {30}