ОДЗ: x>-5, x>5
x∈(5;+∞)
log2(5+x)≥-log2(x-5)
log2(5+x)≥log2(1/(x-5))
x+5≥1/(x-5)
(x²-25-1)/(x-5)≥0
(x²-26)/(x-5)≥0
[-√26;5)∪[√26;+∞), соответственно ОДЗ остается x∈[√26;+∞)
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
Verified answer
ОДЗ: x>-5, x>5
x∈(5;+∞)
log2(5+x)≥-log2(x-5)
log2(5+x)≥log2(1/(x-5))
x+5≥1/(x-5)
(x²-25-1)/(x-5)≥0
(x²-26)/(x-5)≥0
[-√26;5)∪[√26;+∞), соответственно ОДЗ остается x∈[√26;+∞)