Ответ:
Объяснение:
a) x² - 2x - 15 < 0
x² + 3x - 5x - 15 > 0
x(x + 3) - 5(x + 3) > 0
(x + 3)(x - 5) > 0
x + 3 < 0 x - 5 > 0
x < -3 x > 5
x ∈ (-∞, -3) ∪ (5, +∞)
б) -x² + 6x ≥ 0
-x(x - 6) ≥ 0
x(x - 6) ≤ 0
x ≥ 0 x - 6 ≤ 0
x ≤ 6
x ∈ [0, 6]
в) ![\frac{4 - x}{5x - 2} \geq 0 \frac{4 - x}{5x - 2} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7B4%20-%20x%7D%7B5x%20-%202%7D%20%20%5Cgeq%20%200)
4 - x ≤ 0 5x - 2 > 0
x ≤ 4 5x > 2
x > ![\frac{2}{5} \frac{2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B5%7D)
x ∈ (
, 4]
г) x(x - 9)(x + 2) > 0
x < 0 (x - 9)(x + 2) > 0
x - 9 > 0 x + 2 > 0
x > 9 x > -2
x ∈ (-2, 0) ∪ (9, +∞)
д) ![\frac{(x - 3)(3x + 3)}{2x + 5} \geq 0 \frac{(x - 3)(3x + 3)}{2x + 5} \geq 0](https://tex.z-dn.net/?f=%5Cfrac%7B%28x%20-%203%29%283x%20%2B%203%29%7D%7B2x%20%2B%205%7D%20%5Cgeq%200)
2x + 5 > 0 (x - 3)(3x + 3) ≥ 0
2x > -5 x - 3 ≥ 0 3x + 3 ≤ 0
x >
x ≥ 3 3x ≤ -3
x ≤ -1
x ∈ (
, -1] ∪[3, +∞)