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<br>[(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
x₂ = 5³/² = 5√5
1/5 ≤ x ≤ 5√5