log0,7 1/(x-2)(x-3)>=log0,7 3/(3-x)(6-x)

Ответ:

х ∈ (-∞; 0] ∪ (3; +∞)

Пошаговое объяснение:

log0,7 1/(x-2)(x-3) ≥ log0,7 3/(3-x)(6-x)

0 < 1/(x-2)(x-3) ≤ 3/(3-x)(6-x)

1/(x-2)(x-3) > 0

(3-x)(6-x) ≤ 3(x-2)(x-3)

(x-2)(x-3) > 0

3(x-2)(x-3) - (3-x)(6-x) ≥ 0

(x-2)(x-3) > 0

(3x-6)(x-3) + (x-3)(6-x) ≥ 0

(x-2)(x-3) > 0

(x-3)2x ≥ 0

х ∈ (-∞; 2) ∪ (3; +∞)

х ∈ (-∞; 0] ∪ [3; +∞)

х ∈ (-∞; 0] ∪ (3; +∞)