Question

Помогите пожалуйста!)

Answer 1

(2+log(3)0,5)/(3-log(3)x-1≥0
(2+log(3)0,5-3+log(3)x)/(log(3)-3)≤0
(log(3)0,5x-1)/(log(3)x-3)≤0
log(3)0,5x=1⇒0,5x=3⇒x=6
log(3)x=3⇒x=9
x∈[6;9)