(log₂(32x) - 1) / ( log₂²x - log₂x⁵) ≥ -1
одз x>0
log₂²x - log₂x⁵= log₂²x - 5log₂x=log₂x (log₂x-5)≠0
log₂x≠0 x≠1 log₂x-5≠0 x≠32
(log₂(32x) - 1) / ( log₂²x - log₂x⁵) + 1≥0
(log₂(32) + log₂x - 1) / ( log₂²x - 5log₂x) +1≥0
(5 + log₂x - 1 + log₂²x - 5log₂x) / ( log₂²x - 5log₂x)≥0
(log₂²x - 4log₂x+4) / ( log₂x( log₂x-5)) ≥ 0
log₂x=t
(t²-4t+4) / t(t-5) ≥ 0
(t-2)² / t(t-5) ≥ 0
+++++++ 0 -------- 2 --------- 5 +++++++++
t∈(-∞ 0) U {2} U (5 +∞)
log₂x < 0 x∈(0 1)
log₂x=2 x = 4
log₂x> 5 x ∈(32 +∞)
ответ x∈( 0 1) U {2} U (32 +∞)