3^x=a
(13-5a)/(a²-12a+27)-1/2≥0
(26-10a-a²+12a-27)/(a²-12a+27)≥0
(a²-2a+1)/(a²-12a+27)≤0
(a-1)²/(a²-12a+27)≤0
a-1=0⇒a=1
a²-12a+27=0
a1+a2=12 U a1*a2=27⇒a1=3 U a2=9
+ + _ +
--------------[1]------------(3)---------------(9)--------------
a=1 U 33^x=1⇒x=0 U 3<3^x<9⇒1<x<2<br>x∈(1;2) U {1}