1 + (1/2) log (3^1/2, (x+5) / (x+3)) >=log (9, (x+1) ^2)

1. ОДЗ:

{ (x + 5) / (x + 3) > 0;

{ (x + 1) ^2 > 0;

{x ∈ (-∞; - 5) ∪ (-3; ∞);

{x ≠ 0;

x ∈ (-∞; - 5) ∪ (-3; 0) ∪ (0; ∞).

2.

1 + (1/2) log (3^ (1/2), (x + 5) / (x + 3)) ≥ log (9, (x + 1) ^2);

1 + log (3^ (1/2), ((x + 5) / (x + 3)) ^ (1/2)) ≥ log (3^2, (x + 1) ^2);

1 + log (3, (x + 5) / (x + 3)) ≥ log (3, |x + 1|);

log (3, 3 (x + 5) / (x + 3)) ≥ log (3, |x + 1|);

3 (x + 5) / (x + 3) ≥ |x + 1|;

3 (x + 5) / (x + 3) - |x + 1| ≥ 0;

3 (x + 5) / (x + 3) - |x + 1| ≥ 0;

(3 (x + 5) - (x + 3) |x + 1|) / (x + 3) ≥ 0.

3.

1) x ∈ (-∞; - 5) ∪ (-3; - 1);

(3x + 15 + (x + 3) (x + 1)) / (x + 3) ≥ 0;

(3x + 15 + x^2 + 4x + 3) / (x + 3) ≥ 0;

(x^2 + 7x + 18) / (x + 3) ≥ 0;

D = 7^2 - 4 * 18 < 0 - нет корней.

{x ∈ (-∞; - 5) ∪ (-3; - 1);

{x ∈ (-3; ∞);

x ∈ (-3; - 1).

2) x ∈ (-1; 0) ∪ (0; ∞);

(3x + 15 - (x + 3) (x + 1)) / (x + 3) ≥ 0;

(3x + 15 - x^2 - 4x - 3) / (x + 3) ≥ 0;

(-x^2 - x + 12) / (x + 3) ≥ 0;

(x^2 + x - 12) / (x + 3) ≥ 0;

(x + 4) (x - 3) / (x + 3) ≥ 0;

{x ∈ (-4; - 3) ∪ (3; ∞);

{x ∈ (-1; 0) ∪ (0; ∞);

x ∈ (3; ∞).

4. Объединяем:

x ∈ (-3; - 1) ∪ (3; ∞).

Ответ: (-3; - 1) ∪ (3; ∞).