<=>
`( (log_3 3)/(log_3 3+log_3 (x-1)) + (log_3 27)/(2*log_3 (x-1)) ) >=2
<=>
`( 1/ (1+log_3 (x-1)) + 3/(2*log_3 (x-1) ) >=2`
` ( 1/log3 (x-1)+ 3/(2log_3 (x-1) )>=1`
log_3 (x-1) = t
Ограничения?
t<>0
(1/t)+(3/2t)-1>=0
(2+3-2t)/2t>=0
t in (0;5/2)
До этого момента правильно?