11log11 (x2+x-20) ≤12+log11 (x+5) 11/x-4

1. ОДЗ:

{x^2 + x - 20 > 0;

{ (x + 5) ^11 / (x - 4) > 0; { (x + 5) (x - 4) > 0;

{ (x + 5) / (x - 4) > 0; x ∈ (-∞; - 5) ∪ (4; ∞).

2. Преобразуем:

11log11 (x^2 + x - 20) ≤ 12 + log11 ((x + 5) ^11 / (x - 4)); 11log11 ((x + 5) (x - 4)) ≤ 12 + 11log11|x + 5| - 11log|x - 4|; 11log11|x + 5| + 11log11|x - 4| ≤ 12 + 11log11|x + 5| - log11|x - 4|; 11log11|x - 4| + log11|x - 4| ≤ 12; 12log11|x - 4| ≤ 12; log11|x - 4| ≤ 12/12; log11|x - 4| ≤ 1; |x - 4| ≤ 11; {x - 4 ≠ 0;

{-11 ≤ x - 4 ≤ 11; {x ≠ 4;

{-11 + 4 ≤ x ≤ 11 + 4; {x ≠ 4;

{-7 ≤ x ≤ 15; x ∈ [-7; 4) ∪ (4; 15].

3. В итоге:

{x ∈ (-∞; - 5) ∪ (4; ∞);

{x ∈ [-7; 4) ∪ (4; 15]; x ∈ [-7; - 5) ∪ (4; 15].

Ответ: [-7; - 5) ∪ (4; 15].