Verified answer

3log²1/3 (x) -10log1/3 (x) +3≥0

пусть  log1/3 (x)=t ,тогда:

3t²-10t+3≥0

3t²-10t+3=0

{t₁+t₂=10/3,

{t₁*t₂=1.

t₁=1/3

t₂=3

___+___1/3____-____3___+____t

t∈(-∞;1/3]∪[3;+∞)

Обратная замена:

log1/3 (x)∈(-∞;1/3]∪[3;+∞)

x(0;(1/3)^3]∪[(1/3)^(1/3);+∞)

Ответ:x∈(0;1/27]∪[ 1/∛3;+∞)