(x + 1)(x + 4)(x + 8)
──────–– ≥ –1.
(x – 1)(x – 4)(x – 8)
((x+1)(x+4)(x+8)+(x–1)(x–4)(x–8) )/(x–1)(x–4)(x–8) ≥ 0
((x2+5x+4)(x+8)+(x2–5x+4)(x–8) )/(x–1)(x–4)(x–8) ≥ 0
(x3+5x2+4x+8x2+40x+32+x3–5x2+4x–8x2+40x–32 )/(x–1)(x–4)(x–8) ≥ 0
(2x3+88x)/(x–1)(x–4)(x–8) ≥ 0
2х(x2+44)/(x–1)(x–4)(x–8) ≥ 0
x/(x–1)(x–4)(x–8) ≥ 0
Применяем метод интервалов:
_+__ [0] _–_ (1) ___+___ (4) ___–___ (8) ___+___
О т в е т. (– ∞ ;0] U(1;4)U(8;+ ∞ )