Question

Решить это логарифмическое неравенство: ㏒11(8х²+7)-㏒11(х²+х+1)≥㏒11(х/(x+5)+7), 30

Answer 1

$\displaystyle log_{11}(8x^2+7)-log_{11}(x^2+x+1) \geq log_{11}( \frac{x}{x+5}+7)\\\\ODZ: \left \{ {{8x^2+7\ \textgreater \ 0; x^2+x+1\ \textgreater \ 0} \atop { \frac{8x+35}{x+5}\ \textgreater \ 0}} \right.\\\\ \left \{ {{x\in R} \atop { \frac{8x+35}{x+5}\ \textgreater \ 0}} \right.$

___+___ -5___-____-35/8___+____

ОДЗ: (-oo;-5)(-35/8;+oo)

решение:

$\displaystyle log_{11} \frac{8x^2+7}{x^2+x+1} \geq log_{11}( \frac{8x+35}{x+5})\\\\11\ \textgreater \ 1\\\\ \frac{8x^2+7}{x^2+x+1} \geq \frac{8x+35}{x+5}\\\\ \frac{(8x^2+7)(x+5)-(8x+35)(x^2+x+1)}{(x^2+x+1)(x+5)} \geq 0\\\\ \frac{-3x^2-36x}{(x+5)(x^2+x+1)} \geq 0\\\\ \frac{-3x(x+12)}{(x^2+x+1)(x+5)} \geq 0$

____+__-12 _-__-5__+___0___-_____

решением неравенства (-oo;-12] (-5;0]

с учетом ОДЗ

ответ: (-oo;-12] (-35/8;0]