Решите неравенство log0, 5(x^3-3x^2-9x+27)<=log0, 25(x-3) ^4

Question

Решите неравенство log0, 5(x^3-3x^2-9x+27)<=log0, 25(x-3) ^4

RoMaN ShEpElEv Ученик (157), закрыт 1 год назад

log0, 5(x^3-3x^2-9x+27)<=log0, 25(x-3) ^4

Answer 1

log0,5 (x^3-3x^2-9x+27) =< log0,25 (x-3)^4

(x^3-3x^2-9x+27) >= (x-3)^4

x^2*(x-3) - 9*(x-3) >= (x-3)^4

(x-3)(x-3)(x+3) - (x-3)^4 >= 0

(x-3)^2 * [(x+3) - (x-3)^2] >= 0

(x-3)^2 >= 0 при любых х =>

[(x+3) - (x-3)^2] >= 0
x + 3 - x^2 + 6x - 9 >= 0
x^2 - 7x + 6 =< 0
(x-1)(x-6) =< 0
1 =< x =< 6

Answer 2

log0,5(x^3 - 3x^2 - 9x + 27) <= log0,25(x - 3)^4
⇔ 0,5^{\log_{0,5}(x^3 - 3x^2 - 9x + 27)} <= 0,25^{\log_{0,25}(x - 3)^4}
⇔ x^3 - 3x^2 - 9x + 27 <= (x - 3)^4

x^3 - 3x^2 - 9x + 27 <= (x - 3)^4
⇔ x^3 - 3x^2 - 9x + 27 <= x^4 - 12x^3 + 54x^2 - 108x + 81
⇔ x^4 - 13x^3 + 63x^2 - 99x + 54 >= 0

x_1 = 1
x_2 = 2
x_3 = 3
x_4 ≈ 3,38

x < 1 1 < x < 2 2 < x < 3 3 < x < 3,38 x > 3,38
f(x) + - + - +

x ∈ (-∞; 1] ∪ [2; 3) ∪ (3,38; +∞)

Answer 3

log0,5(x^3-3x^2-9x+27) - log0,25(x-3)^4 ≤ 0
log0,5[(x-3)^3 - 18(x-3)] - 4log0,5(x-3) ≤ 0
log2[(x-3)^3 - 18(x-3)] - 4log2(x-3) ≤ 0
log2[(x-3)^3 - 18(x-3)] ≤ 4log2(x-3)
log2[(x-3)^3 - 18(x-3)] ≤ log2(x-3)^4
[(x-3)^3 - 18(x-3)] ≤ (x-3)^4
(x-3)(x-3-3)(x-3+3) - 18(x-3) ≤ (x-3)^4
(x-3)(x^2-6x+9-18) ≤ (x-3)^4
(x-3)(x^2-6x-9) ≤ (x-3)^4
(x-3)(x-3-√9)(x-3+√9) ≤ (x-3)^4
(x-3)(x-3-3)(x-3+3) ≤ (x-3)^4
-9(x-3) ≤ 0
x ≥ 3