Question

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Решите, пожалуйста, со всеми ПОДРОБНОСТЯМИ!

Answer 1

log₂(log₃(log₄(log₅²(133-2x)+7)+25)-1)≤1
область определения 133-2x>0
133>2x
x<133/2=66,5

log₃(log₄(log₅²(133-2x)+7)+25)-1≤2
log₃(log₄(log₅²(133-2x)+7)+25)≤3
log₄(log₅²(133-2x)+7)+25≤27
log₄(log₅²(133-2x)+7)≤2
log₅²(133-2x)+7≤16
log₅²(133-2x)≤9

-3≤log₅(133-2x)≤3
1/125 ≤ 133-2x ≤ 125
1/125 -133≤ -2x ≤ 125-133
(133-1/125)/2 ≥ x ≥ 8/2
66,496 ≥ x ≥ 4

Answer 2

ОДЗ
133-2x>0⇒2x<133⇒⇒
x<66,5
log(3)(log(4)(log(5)²(133-2x)+7)+25)-1≤2
og(3)(log(4)(log(5)²(133-2x)+7)+25)≤3
(log(4)(log(5)²(133-2x)+7)+25≤27
log(4)(log(5)²(133-2x)+7≤2
log(5)²(133-2x)+7≤16
log(5)²(133-2x)-9≤0
(log(5)(133-2x)+3)(log(5)(133-2x)-3)≤0
log(5)(133-2x)=-3 U log(5)(133-2x)=3
-3≤log(5)(133-2x)≤3
1/125≤133-2x≤125
-132,992≤-2x≤-8
4≤x≤66,496
x∈[4;66,496]