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leratrifonova1
@leratrifonova1
August 2022
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помогите пожалуйста решить [tex] \frac{1}{2}^{ x^{2} +2} \ \textgreater \ \frac{1}{2}^{3x} [/tex]
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ShirokovP
Verified answer
(1/2)^(x^2 + 2) > (1/2)^(3x)
Так как 0 <1/2 < 1, тогда знак меняется на противоположный
x^2 + 2 < 3x
x^2 - 3x + 2 < 0
x^2 - 3x + 2 = 0
D = 9 - 8 = 1
x1 =( 3 +1)/2 = 4/2 = 2
x2 = ( 3 -1)/2 = 2/2 = 1
(x - 2)(x - 1) < 0
x ∈ (1; 2)
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Answers & Comments
Verified answer
(1/2)^(x^2 + 2) > (1/2)^(3x)Так как 0 <1/2 < 1, тогда знак меняется на противоположный
x^2 + 2 < 3x
x^2 - 3x + 2 < 0
x^2 - 3x + 2 = 0
D = 9 - 8 = 1
x1 =( 3 +1)/2 = 4/2 = 2
x2 = ( 3 -1)/2 = 2/2 = 1
(x - 2)(x - 1) < 0
x ∈ (1; 2)