Question

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

Answer 1

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

__+____ -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]