5x=t
5x+1=5x·5=5t
25x=(52)x=(5x)2=t2
Неравенство принимает вид
(t/6)+(3t2–10t–25)/(t2–24t–25)–(5/6) ≥ 0.
t2–24t–25=(t+1)(t–25)
D=242–4·(–25)=576+100=676
t=–1 или t=25
(t3–11t2+35t–25)/(6(t+1)(t–25))≥ 0;
(t–5)·(t2–6t+5)/(t+1)(t–25)≥ 0;
t2–6t+5=0
D=36–20=16
t=1 или t=5
(t–5)2(t–1)/(t+1)(t–25)≥ 0.
Применяем метод интервалов.
__–__ (–1) __+__ [1] ___–___ [5] ___–_____ (25) __+__
–1 < t ≤ 1
t > 25
Учитывая, что t=5x и t > 0 при любом х
0 < 5x ≤ 1=50 ⇒ x ≤ 0
5x > 25 ⇒ x > 2
О т в е т. (–∞;0] U (2;+∞)