Question

Решите неравенство ​

Answer 1

log a + log b = log ab

log(2) (2/x - 1) + log(2) (2/x + 1) ≤ log(2) (32x - 1)

одз  2/x - 1 > 0 (2-x)/x > 0   x∈ (0 , 2)

2/x + 1 > 0   (2 + x)/x > 0 x∈(-∞, -2) U (0, +∞)

32x - 1 > 0   x > 1/32

общее x ∈ (1/32, 2)

log(2) (2/x - 1) + log(2) (2/x + 1) ≤ log(2) (32x - 1)

log(2) (2/x - 1)(2/x + 1) ≤ log(2) (32x - 1)

log(2) (4/x² - 1) ≤ log(2) (32x - 1)

4/x² - 1 ≤ 32x - 1

4/x² - 32x ≤ 0

4(1 - 8x³)/x² ≤ 0

4(1 - 2x)(1 + 2x + 4x²)/x² ≤ 0

4(1 + 2x + 4x²)/x² всегда неотрицателен

1 - 2х ≤ 0

х ≥ 1/2

x ∈ [1/2, 2)