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Дима555
@Дима555
August 2021
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log7(x^2 - 4x+5)>log7(2x-3) помогите ,срочно!
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kirichekov
Verified answer
Log₇(x²-4x+5)>log₇(2x-3)
x²-4x+5>0
D<0, x ∈(-∞;∞)
2x-3>0
x>1,5
основание 7>1, знак не меняем
x²-4x+5>2x-3
x²-6x+8>0
x₁=2, x₂=4
+ - +
---------------|------------|--------------- x
2 4
x∈(-∞;2)U(4;∞)
x∈(1,5;∞) ⇒
x∈(1,5;2)U(4;∞)
3 votes
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Answers & Comments
Verified answer
Log₇(x²-4x+5)>log₇(2x-3)x²-4x+5>0
D<0, x ∈(-∞;∞)
2x-3>0
x>1,5
основание 7>1, знак не меняем
x²-4x+5>2x-3
x²-6x+8>0
x₁=2, x₂=4
+ - +
---------------|------------|--------------- x
2 4
x∈(-∞;2)U(4;∞)
x∈(1,5;∞) ⇒
x∈(1,5;2)U(4;∞)