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GinnyS
@GinnyS
July 2022
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log x^2+4x)>_ -1
. 1/5
Срочно!!!!!!!!
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nKrynka
Решение
log₁/₅ (x² + 4x) ≥ - 1
ОДЗ: x² + 4x > 0, x(x + 4) > 0
x = 0 ; x = - 4
x ∈ ( - ∞ ; - 4) (0 ; + ∞)
так как 0 < 1/5 < 1, то
x² + 4x ≤ (1/5)⁻¹
x² + 4x ≤ 5
x² + 4x - 5 ≤ 0
x₁ = - 5
x₂ = 1
x ∈ [- 5 ; 1]
С учётом ОДЗ x∈ [- 5; - 4) ; (0 ; 1]
Ответ:
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Answers & Comments
log₁/₅ (x² + 4x) ≥ - 1
ОДЗ: x² + 4x > 0, x(x + 4) > 0
x = 0 ; x = - 4
x ∈ ( - ∞ ; - 4) (0 ; + ∞)
так как 0 < 1/5 < 1, то
x² + 4x ≤ (1/5)⁻¹
x² + 4x ≤ 5
x² + 4x - 5 ≤ 0
x₁ = - 5
x₂ = 1
x ∈ [- 5 ; 1]
С учётом ОДЗ x∈ [- 5; - 4) ; (0 ; 1]
Ответ: