ОДЗ
{x>0
{2x≠1⇒x≠1/2
{4√x≠1⇒√x≠1/4⇒x≠1/16
x∈(0;1/16) U (1/16;1/2) U (1/2;∞)
Перейдем к основанию 2
6log(2)x/(log(2)x +1)+4(log(2)x + 1)/(log(2)x+4) -1≥0
log(2)x=a
6a/(a+1)+4(a+1)/(a+4)-1≥0
(6a²+24a+4a²+8a+4-a²-4a-a-4)/[(a+1)(a+4)]≥0
(9a²+27a)/[(a+1)(a+4)]≥0
9a(a+3)/[(a+1)(a+4)]≥0
a=0 a=-3 a=-1 a=-4
+ _ + _ +
-----------(-4)----------[-3]---------(-1)----------[0]------------
a<-4⇒log(2)x<-4⇒x<1/16⇒0<x<1/16<br>-3≤a<-1⇒-3≤log(2)x<-1⇒1/8≤x<1/2<br>a≥0⇒log(2)x≥0⇒x≥1
x∈(0;1/16) U [1/8;1/2) U [1;∞)