Home
О нас
Products
Services
Регистрация
Войти
Поиск
igorf5
@igorf5
July 2022
2
3
Report
Помогите решить логарифмическое уравнение
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
wejde
Verified answer
2log₃(x-2)+log₃(x-4)²=0
D(y): x-2>0, x>2
(x-4)²>0, x≠4
x∈(2;4)U(4;+∞)
2log₃(x-2)+2log₃|x-4|=0/:2, x≠4
log₃(x-2)|x-4|=0
log₃(x-2)|x-4|=log₃1
(x-2)|x-4|=1
1) x-4≥0, x≥4, учитывая D(y) → x>4
(x-2)(x-4)=1
x²-4x-2x+8=1
x²-6x+7=0
D=(-6)²-4*1*7=36-28=8
√D=√(4*2)=2√2
x1=(6+2√2)/2=
3+√2
x2=(6-2√2)/2=3-√2 - пост. корень
2) x-4<0, x<4, учитывая D(y) → x∈(2;4)
(x-2)(4-x)=1
(x-2)(x-4)=-1
x²-4x-2x+8=-1
x²-6x+9=0
x²-2*x*3+3²=0
(x-3)²=0
x=3
Ответ:
x=3, x=
3+√2
0 votes
Thanks 0
sedinalana
ОДЗ
{x-2>0⇒x>2
x-4≠0⇒x≠4
x∈(2;4) U (4;∞)
log_3(x-2)²+log_3(x-4)²=0
log_3[(x-2)(x-4)]²=0
[(x-2)(x-4)]²=1
(x-2)(x-4)=-1 U (x-2)(x-4)=1
x²-6x+9=0 U x²-6x+7=0
(x-3)²=0⇒x-3=0⇒x=3
x²-6x+7=0
D=36-28=8
x1=(6-2√2)/2=3-√2∉ОДЗ
x2=3+√2
Ответ {3;3+√2}
2 votes
Thanks 2
igorf5
Почему корень 3-sqrt(2) не удовлетворяет x>2, x(не=4)
sedinalana
3-1,4=1,6, а ОДЗ x∈(2;4) U (4;∞)
More Questions From This User
See All
igorf5
August 2022 | 0 Ответы
kak reshit kubicheskoe uravnenie x328x26x0
Answer
igorf5
August 2022 | 0 Ответы
reshite neravenstvo pozhalujsta664a2319c4622b7883ae154daa9a7b40 60272
Answer
igorf5
July 2022 | 0 Ответы
pomogite reshit uravnenie00b177f7a9bbeaa86102b085702039d7 17034
Answer
igorf5
July 2022 | 0 Ответы
pomogite reshit sistemu s dvumya peremennymi
Answer
igorf5
July 2022 | 0 Ответы
podskazhite kak reshit zadachu po geometrii kak najti obyom kryshi saraya
Answer
igorf5
July 2022 | 0 Ответы
pomogite reshit sistemu s dvumya neizvestnymi pozhalujsta
Answer
igorf5
July 2022 | 0 Ответы
pomogite reshit pokazatelnoe uravnenie kakim metodom ono reshaetsya
Answer
igorf5
July 2022 | 0 Ответы
kakim metodom reshat etu sistemu s dvumya neizvestnymi
Answer
igorf5
July 2022 | 0 Ответы
pomogite reshite pokazatelnoe uravnenie s modulem
Answer
igorf5
July 2022 | 0 Ответы
podskazhite kak reshat zadachi kogda dana obshaya messa dvuh sushestv i obshij obyom
Answer
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 "Помогите решить логарифмическое уравнение..."
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
2log₃(x-2)+log₃(x-4)²=0D(y): x-2>0, x>2
(x-4)²>0, x≠4
x∈(2;4)U(4;+∞)
2log₃(x-2)+2log₃|x-4|=0/:2, x≠4
log₃(x-2)|x-4|=0
log₃(x-2)|x-4|=log₃1
(x-2)|x-4|=1
1) x-4≥0, x≥4, учитывая D(y) → x>4
(x-2)(x-4)=1
x²-4x-2x+8=1
x²-6x+7=0
D=(-6)²-4*1*7=36-28=8
√D=√(4*2)=2√2
x1=(6+2√2)/2=3+√2
x2=(6-2√2)/2=3-√2 - пост. корень
2) x-4<0, x<4, учитывая D(y) → x∈(2;4)
(x-2)(4-x)=1
(x-2)(x-4)=-1
x²-4x-2x+8=-1
x²-6x+9=0
x²-2*x*3+3²=0
(x-3)²=0
x=3
Ответ: x=3, x=3+√2
{x-2>0⇒x>2
x-4≠0⇒x≠4
x∈(2;4) U (4;∞)
log_3(x-2)²+log_3(x-4)²=0
log_3[(x-2)(x-4)]²=0
[(x-2)(x-4)]²=1
(x-2)(x-4)=-1 U (x-2)(x-4)=1
x²-6x+9=0 U x²-6x+7=0
(x-3)²=0⇒x-3=0⇒x=3
x²-6x+7=0
D=36-28=8
x1=(6-2√2)/2=3-√2∉ОДЗ
x2=3+√2
Ответ {3;3+√2}