2x=t
Неравенство примет вид:
2t√(2t–1)/(t–15) ≤ √(2t–1)/(t–8)
или
(2t2–16t–t+15)√(2t–1)/(t–15)(t–8) ≤ 0;
(t–1)(2t–15)√(2t–1)/(t–15)(t–8) ≤ 0.
ОДЗ: 2t–1 ≥ 0.
Решаем методом интервалов на [1/2; + ∞)
[1/2] _+_ [1] __–__[15/2] _+_ (8) __–___ (15)___+__
t=(1/2) или 1 ≤ t ≤ (15/2)
или 8 < t < 15.
Обратная замена
2x=2–1 или 20 ≤ 2x ≤ 2log215/2
или 23 < 2x < 2log215.
О т в е т. {–1} U [0; log215/2]; (3; log215)