Выделяем целую часть у дроби слева.

Делим многочлена x4–5x3+3x–25 на x2–5x ''уголком''

x4–5x3+3x–25 | x2–5x

x4–5x3

––––––––

Неравенство примет вид:

x2+(3x–25)/(x2–5x) ≥ х2–(1/(x–4))+(5/x);

(3x–25)/x·(x–5)+(1/(x–4))–(5/x)≥ 0;

((3x–25)·(x–4)+(x2–5x)–5·(x–5)8(x–4))/(x·(x–4)·(x–5))≥ 0;

(–x2+3x)/(x·(x–4)·(x–5))≥ 0;

или

(х–3))/((x–4)·(x–5))≤ 0 при х≠0.

_–_0 _–_ [3] _+_ (4) _–__ (5) _+__

О т в е т. (–∞;0)U(0;3]U(4;5)