3^x + 270/3^x >= 37 |*3^x (всегда 3^x > 0)
3^x = t
t^2 - 37t + 270 >= 0
по Виету
t1*t2 = 270
t1 + t2 = 37
(t - 10)(t - 27) >=0
t ∈ (-∞, 10] U [27, +∞)
3^x < = 10
x <= log(3) 10
3^x >= 27
x >= 3
x ∈ (-∞, log(3) 10] U [3, +∞) это решение первого
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log(x+1) (x+3)/6 <= 0
одз x + 1 > 0 x > -1
x + 1 ≠ 1 x ≠ 0
x + 3 > 0 x>-3
x∈ (-1, +∞ ) ∩ {0}
log(x+1) (x+3)/6 <= log(x+1) 1 ⇔ (x + 1 -1)((x+3)/6 - 1) <=0
x(x - 3)/6 <=0
+++++++[0] --------------- [3] +++++++++
x ∈ [0,3] пересекаем с решением первого x ∈ (-∞, log(3) 10] U [3, +∞) и одз x∈ (-1, +∞ ) ∩ {0}
Ответ x ∈(0, log(3) 10] U {3}
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Verified answer
3^x + 270/3^x >= 37 |*3^x (всегда 3^x > 0)
3^x = t
t^2 - 37t + 270 >= 0
по Виету
t1*t2 = 270
t1 + t2 = 37
(t - 10)(t - 27) >=0
t ∈ (-∞, 10] U [27, +∞)
3^x < = 10
x <= log(3) 10
3^x >= 27
x >= 3
x ∈ (-∞, log(3) 10] U [3, +∞) это решение первого
---
log(x+1) (x+3)/6 <= 0
одз x + 1 > 0 x > -1
x + 1 ≠ 1 x ≠ 0
x + 3 > 0 x>-3
x∈ (-1, +∞ ) ∩ {0}
log(x+1) (x+3)/6 <= log(x+1) 1 ⇔ (x + 1 -1)((x+3)/6 - 1) <=0
x(x - 3)/6 <=0
+++++++[0] --------------- [3] +++++++++
x ∈ [0,3] пересекаем с решением первого x ∈ (-∞, log(3) 10] U [3, +∞) и одз x∈ (-1, +∞ ) ∩ {0}
Ответ x ∈(0, log(3) 10] U {3}