Решить два логарифмических неравенства

Решение 1)  2log₁/₂ (x - 2) + log₂ (x² - 2x - 1) < 1  - 2log₂ (x - 2) + log₂ (x² - 2x - 1) < 1 - 2log₂ (x - 2)⁻² + log₂ (x² - 2x - 1) < 1 log₂ [(x² - 2x - 1)/(x - 2)²] < log₂ 2 так как 2 > 1, то (x² - 2x - 1)/(x - 2)² < 2 [(x² - 2x - 1) - 2*(x - 2)²]/(x - 2)² < 0 [x² - 2x - 1 - 2*(x² - 4x + 4)]/(x - 2)² < 0 [x² - 2x - 1 - 2x² + 8x - 8)]/(x - 2)² < 0 (- x² +6x - 9)/(x - 2)² < 0 (x² - 6x + 9)/(x - 2)² > 0 (x - 3)² / (x - 2)² > 0 x∈ (- ∞ ; 2)∪(2 ; 3)∪(- (3 ; + ∞) 2)  2log₅ x - (log₅ 125 / log₅ x) ≤ 1 ОДЗ: x > 0 2log₅ x - (log₅ 5³/ log₅ x) ≤ 1 2log₅ x - (3*log₅ 5/ log₅ x) ≤ 1 2log²₅ x - log₅ x - 3 ≤ 0 log₅ x = t 2t² - t - 3 = 0 D = 1 + 4*2*3 = 25 t₁ = (1 - 5)/4 = - 1 t₂ = (1 + 5)/4 = 3/2 1)  log₅ x = - 1 x = 5 ⁻¹ x₁ = 1/5 2)  log₅ x = 3/2 x ₂ = 5³/² = 5√5 1/5 ≤ x ≤ 5√5