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alla3
@alla3
July 2022
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Решить неравенство:
log2(x^2 - 3x + 2) ≤ 1 + log2(x-2)
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ShirokovP
Verified answer
ОДЗ
x > 2
log2(x^2 - 3x + 2) ≤ log2(2) + log2(x - 2)
log2(x^2 - 3x + 2) ≤ log2 (2x - 4)
x^2 - 3x + 2 ≤ 2x - 4
x^2 - 5x + 6 ≤ 0
x^2 - 5x + 6 = 0
D = 25 - 24 = 1
x1 = (5 + 1)/2 = 3
x2 = (5 - 1)/2 = 2
(x - 3)(x - 2) ≤ 0
x ∈ [ 2; 3]
+ ОДЗ
x ∈ (2; 3]
5 votes
Thanks 3
alla3
Спасибо вам, добрый человек!
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Answers & Comments
Verified answer
ОДЗx > 2
log2(x^2 - 3x + 2) ≤ log2(2) + log2(x - 2)
log2(x^2 - 3x + 2) ≤ log2 (2x - 4)
x^2 - 3x + 2 ≤ 2x - 4
x^2 - 5x + 6 ≤ 0
x^2 - 5x + 6 = 0
D = 25 - 24 = 1
x1 = (5 + 1)/2 = 3
x2 = (5 - 1)/2 = 2
(x - 3)(x - 2) ≤ 0
x ∈ [ 2; 3]
+ ОДЗ
x ∈ (2; 3]