5x > 0 ⇒ 5x ≠ 0
((1/4)·(4/5)x)/(1–(4/5)x) – (1/(2+4·(4/5)x)) ≥ 0
Замена переменной
(4/5)x=t
t > 0
(t/(4–4t)) – (1/(2+4t)) ≥ 0
(2t2+3t–2)/(4·(1–t)·(1+2t)) ≥ 0
(2t+1)(t+2)/(t–1)(2t+1) ≤ 0
C учетом t > 0 получим
0 < t < 1
0 < (4/5)x < 1 ⇒ x > 0