Методом интервалов Благодарю

log ab = log a + log b

(log(5) (125x)*log(3) (81x)) / (x² - |x|) ≤ 0

одз 125x > 0 x > 0

81x > 0  x > 0

x² - |x| ≠ 0   x ≠ 0  x ≠ -1 x ≠ 1

x∈(0, 1) U (1, +∞)

так как x > 0 модуль снимаем как х

((log(5) (125) + log(5) x)*(log(3) (81) + log(3) x) / x(x - 1) ≤ 0

81 = 3^4

125 = 5^3

(3 + log(5) x)(4 + log(3) x) / x(x - 1) ≤ 0

log(5) x = -3    x = 5^(-3) = 1/125

log(3) x = -4    x = 3^(-4) = 1/81

(0) [1/125][1/81](1)

x ∈ (0, 1/125] U [ 1/81, 1)