(127-7^x-7^4-x I 20-7^x I ) / 29 - I 20 - 7^x больше или равно 1. Created by nscales. matematika-ru

29 - I 20 - 7^x больше или равно 1...

July 2022 1 4 Report

(127-7^x-7^4-x I 20-7^x I ) / 29 - I 20 - 7^x больше или равно 1

Answers & Comments

konik1 (127-7^x-7^(4-x)*|20-7^x|) / (29 - |20 - 7^x|) ≥ 1   y=7^x
(127-y-7^4/y *|20-y|) / (29 - |20 - y|) ≥ 1
127-y-7^4/y *|20-y| ≥ 29 - |20 - y|
127-y-29 ≥ 7^4/y *|20-y| - |20 - y|
127-y-29 ≥ |20-y|*(7^4/y - 1)
(127-y-29)/(7^4/y - 1) ≥ |20-y| Раскрываем модуль и решаем:

(98-y)/(7^4/y - 1) ≥ 20-y
-(98-y)/(7^4/y - 1) ≤ 20-y

(98-y)/(7^4/y - 1) ≥ 20-y
98-y≥ (20-y)*(7^4/y - 1)
98y-y^2 ≥ (20-y)*(7^4 - y)
98y-y^2 ≥ 20*7^4-20y-(7^4)y+y^2
98y-y^2 ≥ 48020-20y-2401y+y^2
0 ≥ 48020-2519y+2y^2 найдём корни квадратного уравнения ≈19.361 и ≈1240.1
2(y-19.361)(y-1240.1) ≤ 0
y ≥ 19.361 7^x≥19.361   x ≥ (log19.361 )/log7 x ≥ 1,5228
y ≤ 1240.1 7^x≤1240.1   x ≤ (log1240.1)/log7 x ≤ 3,66
логорифмы по любым, но одинаковым основаниям, вычисляются на калькуляторе.
-(98-y)/(7^4/y - 1) ≤ 20-y
(y-98)/(7^4/y - 1) ≤ 20-y
y-98 ≤ (20-y)*(7^4/y - 1)
y^2-98y ≤ (20-y)*(2401 - y)
y^2-98y ≤ 48020- 20y-2401y+y^2
-98y ≤ 48020-2421y
2323y ≤ 48020
y ≤ 48020/2323 у ≈ 20,671
y ≤ 20,671 ; 19,361 минимальное значение у
19.361 ≤ 7^x ≤ 1240.1
(log19.361)/log7≤ x ≤ (log1240.1)/log7
1,5228 ≤ x ≤ 3,66 ответ

0 votes Thanks 0

Answers & Comments

5 najdite v sm perimetr etogo treugolnika pomogite pliz

kvadratnoe uravnenie korni kotorogo na 3 edinicy bolshe kornej uravneniya x23x 3

1 tg 2 x2cos2x

logx log4 16 x 240 bolshe ili ravno1

izvestno chto h1 i h 2 korni uravneniya h 2 9 h a 0 h 3 i h4 korni

6 ishodnogo kolichestva deneg polozhili vo vtoroj bank a ostavshuyusya chast v perv

3 yavlyaetsya ego resheniem

kak reshit najdite tg a esli vypolnyaetsya ravenstvo 10tg a 5 tg a sin a

рекомендуемые вопросы

Helpful Links

Helpful Social