Home
О нас
Products
Services
Регистрация
Войти
Поиск
NNNgayurka
@NNNgayurka
December 2021
1
14
Report
Допоможіть розв'язати будь-ласка 2.6; 2.8; 2.12; 2.14; Даю 90 балів
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
dasdasfa
2.6 7^((2x^2-5x-9)/2)=(√2 ^3log(2) 7
7^((2x^2-5x-9)/2) =2^log(2) (7^ 3/2)
7^((2x^2-5x-9)/2)=7^(3/2)
(2x^2 -5x -9)/2=3/2
2x^2-5x-9=3; 2x^2-5x-12=0; D=25-4*2*(-12)=25+96=121=11^2;
x1=(5-11)/4=-6/4=-3/2; x2=4
Ответ. -1,5; 4
2.8 5*(5^ (-2))^sin^2 x+ 4* 5^cos2x =5^2)^(1/2 sin2x;
5^(1-2sin^2 (x) +4*5^cos(2x) =5^sin(2x);
5^(1-1+cos(2x) +4*5^cos(2x) =5^sin(2x)
5*5^cos(2x)=5^sin(2x)
5^cos(2x) =5^sin(2x); cos(2x)=sin2x; cos(2x)-sin(2x)=0; -tg(2x)=-1
tg(2x)=1; 2x=π/4+πn
x=π/8+π/2 *n
---------------------
2.12 2^(3(x-3)/(3x-7)) * ∛√(1/4)^(3x-1)/(x-1)=1 6
2^(3(x-3)/(3x-7)) * (2^-2)^((3x-1)/(6(x-1))=1; ∛√a=√a=a^(1/6)
2^(3(x-3)/(3x-7)) *2^(-2*(3x-1 /(6(x-1))=1; 2^0=1
(3x-9)/(3x-7)+((1-3x)/(3x-3)=0
((3x-9)*(3x-3) +(1-3x) *(3x-7) ) /((3x-7)(3x-3)=0
9x^2 -9x-27x+27 +3x-7-9x^2 +21x)=0; (3x-7)(3x-3)≠0
-12x+20=0; x=20/12; x=5/3
0 votes
Thanks 0
More Questions From This User
See All
NNNgayurka
August 2022 | 0 Ответы
dam za pravilnoe reshenie 99 ballov pri izohornom ohlazhdenii 32 g kisloroda vzya
Answer
NNNgayurka
August 2022 | 0 Ответы
dam za pravilnoe reshenie 99 balov v vertikalnom cilindre pod porshnem massoj 1
Answer
NNNgayurka
October 2021 | 0 Ответы
ukr v vagone dvizhushegosya pryamolinejno s uskoreniem a k potolku podveshen sharik
Answer
NNNgayurka
September 2021 | 0 Ответы
s2 za kakoe vremya ot nachala dvizheniya normalnoe uskorenie tochki budet1 ravno
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 "Допоможіть розв'язати будь-ласка 2.6; 2.8; 2.12; 2.14; Даю 90 балів..."
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
7^((2x^2-5x-9)/2) =2^log(2) (7^ 3/2)
7^((2x^2-5x-9)/2)=7^(3/2)
(2x^2 -5x -9)/2=3/2
2x^2-5x-9=3; 2x^2-5x-12=0; D=25-4*2*(-12)=25+96=121=11^2;
x1=(5-11)/4=-6/4=-3/2; x2=4
Ответ. -1,5; 4
2.8 5*(5^ (-2))^sin^2 x+ 4* 5^cos2x =5^2)^(1/2 sin2x;
5^(1-2sin^2 (x) +4*5^cos(2x) =5^sin(2x);
5^(1-1+cos(2x) +4*5^cos(2x) =5^sin(2x)
5*5^cos(2x)=5^sin(2x)
5^cos(2x) =5^sin(2x); cos(2x)=sin2x; cos(2x)-sin(2x)=0; -tg(2x)=-1
tg(2x)=1; 2x=π/4+πn
x=π/8+π/2 *n
---------------------
2.12 2^(3(x-3)/(3x-7)) * ∛√(1/4)^(3x-1)/(x-1)=1 6
2^(3(x-3)/(3x-7)) * (2^-2)^((3x-1)/(6(x-1))=1; ∛√a=√a=a^(1/6)
2^(3(x-3)/(3x-7)) *2^(-2*(3x-1 /(6(x-1))=1; 2^0=1
(3x-9)/(3x-7)+((1-3x)/(3x-3)=0
((3x-9)*(3x-3) +(1-3x) *(3x-7) ) /((3x-7)(3x-3)=0
9x^2 -9x-27x+27 +3x-7-9x^2 +21x)=0; (3x-7)(3x-3)≠0
-12x+20=0; x=20/12; x=5/3