Question

log2(17x^2+16)-log2(x^2+x+1)>=log2((x/x+10)+16)

Answer 1

$\displaystyle log_2(17x^2+16)-log_2(x^2+x+1) \geq log_2( \frac{x}{x+10}+16)\\\\ODZ: \left \{ {{17x^2+16\ \textgreater \ 0; x^2+x+1\ \textgreater \ 0} \atop { \frac{x+16x+160}{x+10}\ \textgreater \ 0}} \right.\\\\ \left \{ {{x\in R} \atop { \frac{17x+160}{x+10}\ \textgreater \ 0} \right.$

__+____ -10__-____ -160/17___+_____

ODZ: (-oo;-10)(-160/17;+oo)

решение

$\displaystyle log_2( \frac{17x^2+16}{x^2+x+1}) \geq log_2( \frac{17x+160}{x+10})\\\\ 2\ \textgreater \ 1\\\\ \frac{17x^2+16}{x^2+x+1} \geq \frac{17x+160}{x+10}$

$\displaystyle \frac{(17x^2+16)(x+10)-(17x+160)(x^2+x+1)}{(x^2+x+1)(x+10)} \geq 0\\\\ \frac{-7x^2-161x}{(x^2+x+1)(x+10)} \geq 0\\\\ \frac{-x(7x+161)}{(x^2+x+1)(x+10)} \geq 0$

__+___-23__-___-10___+___0____-_

ответов в неравенстве (-oo;-23] (-10;0]

с учетом ОДЗ

ОТВЕТ: (-oo;-23] (-160/17; 0]