au456

Verified answer

ОДЗ  x>0⇒x∈(0;∞)(log(4)x+3)/(log(4)x-3)+(log(4)x-3)/(log(4)x+3)--(4log(4)x+16)/[(log(4)x-3)(log(4)x+3)]≥0log(4)x=a((a+3)/(a-3)+(a-3)/(a+3)-(4a+16)/[(a-3)(a+3)]≥0[(a+3)²+(a-3)²-(4a+16)]/([(a-3)(a+3)]≥0(a²+6a+9+a²-6a+9-4a-16)/[(a-3)(a+3)]≥0(2a²-4a+2)/[(a-3)(a+3)]≥02(a²-2a+1)/[(a-3)(a+3)]≥02(a-1)²/[(a-3)(a+3)]≥0a-1=0⇒a=1a-3=0⇒a=3a+3=0⇒a=-3                   +                      _                      _                     +----------------------(-3)-----------------[1]-------------------(3)---------------------a<-3⇒log(4)x<-3⇒x<1/64 +ОДЗ⇒0<x<1/64a=1⇒log(4)x=1⇒x=4a>3⇒log(4)x>3⇒>x>64Ответ x∈(0;1/64) U (64;∞) U {4}