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unique2
@unique2
August 2022
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Помогите решить пожалуйста!!!!!!!!!!!!!!!!!
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sedinalana
Verified answer
[(3/5)²]^[(x-3)/2x]≥√(3/5)
(3/5)^[(x-3)/x]≥√(3/5)
(x-3)/x≤1/2
(x-3)/x-1/2≤0
(2x-6-x)/(2x)≤0
(x-6)/x≤0
x=6 x=0
+ _ +
----------------(0)---------------------[6]---------------
x∈(0;6]
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Answers & Comments
Verified answer
[(3/5)²]^[(x-3)/2x]≥√(3/5)(3/5)^[(x-3)/x]≥√(3/5)
(x-3)/x≤1/2
(x-3)/x-1/2≤0
(2x-6-x)/(2x)≤0
(x-6)/x≤0
x=6 x=0
+ _ +
----------------(0)---------------------[6]---------------
x∈(0;6]