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youonelove
@youonelove
July 2022
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Показательные неравенства. Помогите решить, пожалуйста.
[tex]( (\frac{3}{7} )^{ \frac{1}{ x^{2} } } )^{ x^{2} -2x} \geq 1[/tex]
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Answers & Comments
MizoriesKun
(3/7)^(1/x²)^(x²-2x)≥1 ОДЗ: x≠0
(3/7)^((x²-2x)/x²)≥1
(7/3)^((2x-x²)/x²)≥(7/3)^0 ⇒
(2x-x²)/x²≥0
x²>0 ⇒
2x-x²≤0
x*(2-x)≤0
-∞_________+________0________-_________2________+________+∞
x∈(0;2].
0 votes
Thanks 1
sangers1959
Verified answer
(3/7)^1/x² =(3/7)^x⁻²
--------------------------------
((3/7)^x⁻²)^(x²-2x) ≥ 1
((3/7)^(x²⁻²-2x¹⁻² ) ≥ 1
((3/7)^(x⁰ -2x⁻¹ ) ≥ 1
((3/7)^(1- 2x⁻¹ ) ≥ 1 т.к 3/7<1 1-2х⁻¹≤0
1-2/х≤0
х-2≤0
х≤ 2 х∈(-∞; 2]
0 votes
Thanks 0
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Answers & Comments
(3/7)^((x²-2x)/x²)≥1
(7/3)^((2x-x²)/x²)≥(7/3)^0 ⇒
(2x-x²)/x²≥0
x²>0 ⇒
2x-x²≤0
x*(2-x)≤0
-∞_________+________0________-_________2________+________+∞
x∈(0;2].
Verified answer
(3/7)^1/x² =(3/7)^x⁻²--------------------------------
((3/7)^x⁻²)^(x²-2x) ≥ 1
((3/7)^(x²⁻²-2x¹⁻² ) ≥ 1
((3/7)^(x⁰ -2x⁻¹ ) ≥ 1
((3/7)^(1- 2x⁻¹ ) ≥ 1 т.к 3/7<1 1-2х⁻¹≤0
1-2/х≤0
х-2≤0
х≤ 2 х∈(-∞; 2]