7) Решить неравенство: 1) log (x-3)+log (9-x)≥−3 , 2 2 2) log2x-3log2 x ≤ 4.

Разбираемся:

1) log₂(x - 3) + log₂(9 - x) ≥ -3 => log₂((x - 3)(9 - x)) ≥ -3 => (x - 3)(9 - x) ≥ 2^(-3) => (x - 3)(9 - x) ≥ 1/8 => -x² + 12x - 27 ≥ 1/8 => -8x² + 96x - 216 ≥ 1 => 8x² - 96x + 217 ≤ 0. D = 96² - 4 * 8 * 217 = 9216 - 6944 = 2272. x1,2 = (96 ± √(2272)) / 16

Неравенство: x - 3 > 0 => x > 3; 9 - x > 0 => x < 9. Следовательно, 3 < x < 9

2) (log₂(x))² - 3log₂(x) ≤ 4 => (log₂(x))² - 3log₂(x) - 4 ≤ 0. Пусть t = log₂(x). t² - 3t - 4 ≤ 0. (t - 4)(t + 1) ≤ 0. -1 ≤ t ≤ 4. -1 ≤ log₂(x) ≤ 4 => 2^(-1) ≤ x ≤ 2^4 => 1/2 ≤ x ≤ 16