log 1 2 6-x≤-2 x+1

log_1/2((6-x) / (x+1)) ≤ -2

ОДЗ: (6-x) / (x+1) > 0

Метод интервалов: x ∈ (-1; 6)

(6-x) / (x+1) ≥ (1/2)^-2

(6-x) / (x+1) ≥ 4

(6-x) / (x+1) - 4 ≥ 0

(6-x - 4(x+1)) / (x+1) ≥ 0

(6-x - 4x - 4) / (x+1) ≥ 0

(-5x + 2) / (x+1) ≥ 0

(5x - 2) / (x+1) ≤ 0

x ∈ (-1; 2/5]

Ответ: (-1; 0.4]