Решить два логарифмических неравенства

Решение1)  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 2так 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) ≤ 1ОДЗ: x > 02log₅ x - (log₅ 5³/ log₅ x) ≤ 1 2log₅ x - (3*log₅ 5/ log₅ x) ≤ 12log²₅ x - log₅ x - 3 ≤ 0log₅ x = t 2t² - t - 3 = 0D = 1 + 4*2*3 = 25t₁ = (1 - 5)/4 = - 1t₂ = (1 + 5)/4 = 3/21)  log₅ x = - 1x = 5 ⁻¹x₁ = 1/52)  log₅ x = 3/2x ₂ = 5³/² = 5√51/5 ≤ x ≤ 5√5