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alla3
@alla3
July 2022
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Решить неравенство:
log2(x^2 - 3x + 2) ≤ log2(x-2) + 1
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sangers1959
Verified answer
Log₂(x²-3x+2)≤log₂(2*(x-2))+1
ОДЗ: x²-3x+2>0 x²-3+2=0 D=1 x₁=2 x₂=1 (x-2)(x-1)>0 x∈(-∞;1)U(2;+∞)
2*(x-2)>0 x>2 ⇒ x∈(2;+∞).
log₂(x²-3x+2)≤log₂(2*(x-2))+log₂2
log₂(x²-3x+2)≤log₂(4*(x-2))
x²-3x+2≤4x-8
x²-7x+10≤0
x²-7x+10=0 D=9
x₁=5 x₂=2
(x-5)(x-2)≤0
-∞______+______2_______-______5______+_______+∞
x∈[2;5].
Учитывая ОДЗ: x∈(2;5).
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Answers & Comments
Verified answer
Log₂(x²-3x+2)≤log₂(2*(x-2))+1ОДЗ: x²-3x+2>0 x²-3+2=0 D=1 x₁=2 x₂=1 (x-2)(x-1)>0 x∈(-∞;1)U(2;+∞)
2*(x-2)>0 x>2 ⇒ x∈(2;+∞).
log₂(x²-3x+2)≤log₂(2*(x-2))+log₂2
log₂(x²-3x+2)≤log₂(4*(x-2))
x²-3x+2≤4x-8
x²-7x+10≤0
x²-7x+10=0 D=9
x₁=5 x₂=2
(x-5)(x-2)≤0
-∞______+______2_______-______5______+_______+∞
x∈[2;5].
Учитывая ОДЗ: x∈(2;5).