Question

ПОМОГИТЕ ПОЖАЛУЙСТА, ОЧЕНЬ СРОЧНО

Answer 1

$\displaystyle \bigg(1 \frac{9}{16}\bigg)^{log_7(x+1)}\ \textless \ \bigg( \frac{4}{5}\bigg)^{log_{1/7}(x+3)} \\\\ODZ: x+1\ \textgreater \ 0; x+3\ \textgreater \ 0\\\\x\ \textgreater \ -1$

$\displaystyle \bigg( \frac{25}{16}\bigg)^{log_7(x+1)}\ \textless \ \bigg(\frac{5}{4}\bigg)^{log_7(x+3)}$

Основание больше 1

$\displaystyle \bigg( \frac{5}{4}\bigg)^{2log_7(x+1)}\ \textless \ \bigg( \frac{5}{4}\bigg)^{log_7(x+3)}\\\\2log_7(x+1)\ \textless \ log_7(x+3)\\\\(x+1)^2\ \textless \ x+3\\\\x^2+2x+1-x-3\ \textless \ 0\\\\x^2+x-2\ \textless \ 0\\\\x_1=1; x_2=-2$

__+___ - 2 __-___ 1 ____+____ (-2;1)

и ОДЗ
__________-1 ___+___________ x>-1

ОТВЕТ (-1;1)