Предмет: Алгебра, автор: Yulya06

Помогите пожалуйста решить! Очень нужно!

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Автор ответа: MizoriesKun

0

㏒₁₎₂(х-5)> -4㏒₁₎₃⁴√(1/3) ОДЗ х-5 >0 x>5

㏒₁₎₂(х-5)> ㏒₁₎₃(⁴√(1/3))⁻⁴

㏒₁₎₂(х-5)> ㏒₁₎₃ 3

㏒₁₎₂(х-5)> -1

х-5 < (1/2)⁻¹

x-5<2

x<7 c учетом ОДЗ х∈(5;7)

(5 4/9)^log₅x > (3/7)^log₁₎₅(5x-6) ОДЗ х>0 ,5x-6>0 x>6/5 (1 1/5)

(49/9)^log₅x > (3/7)^ -log₅(5x-6)

(7/3)^2log₅x > (7/3)^log₅(5x-6)

log₅x² > log₅(5x-6)

x²>5x-6

x²-5x+6>0

D=25-24=1

х=(5+1)/2=3

х=(5-1)/2=2

с учетом ОДЗ х∈(1 1/5 ;2 )∪( 3;+∞)

Автор ответа: MizoriesKun

0

в последнем промежутки не написала (( сижу думаю

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