Question

Решите неравенство.
СПАСАЙТЕ!!!

Answer 1

$log_ frac{1}{2} |1-x|-log_{x-1}2 leq 2 \ O.D.3.: begin{cases} x-1 textgreater 0 \ x-1 neq 1 \ |1-x| textgreater 0 end{cases} = textgreater begin{cases} x textgreater 1 \ x neq 2 end{cases}$
$log_ {2} |1-x|+frac{1}{log_2(x-1)} +2 geq 0 \ Pi pu x textgreater 1 |1-x|=x-1 = textgreater log_ {2} (x-1)+frac{1}{log_2(x-1)} +2 geq 0 \ 3AMEHA: log_2(x-1)=t$
$t+ frac{1}{t} +2 geq 0 \ frac{t^2+2t+1}{t} geq 0 \ frac{(t+1)^2}{t} geq 0$
  -      -      +
----|------owwww->           ⇒ t ∈{-1}∪(0;+∞)
   -1       0
$1) log_2(x-1)=-1 = textgreater x-1=0,5 = textgreater x=1,5; \ 2) log_2(x-1) textgreater 0 = textgreater x-1 textgreater 1 = textgreater x textgreater 2.$
C учетом неравенств в ОДЗ получим ответ: х∈{1,5}∪(2; +∞).