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Tomoki
@Tomoki
August 2021
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Помогите решить пожалуйста. :)
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LFP
Verified answer
:::::::::::::::::::::::::::::::::::::::
1 votes
Thanks 0
oganesbagoyan
Verified answer
Task/26450423
-------------------
решить неравенство
3.
2 ^(3/(1- x) ) ≤ 0,5 ^(1/(3x+1) )
2 ^(3/(1- x)) ≤ 2 ^(-1/(3x+1) ; т.к. основания 2 > 1 ,то
3
/
(1- x) ≤ -1
/
(3x+1) ;
3
/
(x -1) - 1
/
(3x+1) ≥ 0 ;
( 3(3x+1) -(x - 1) )
/
(x -1)(3x+1) ≥ 0 ;
8(x +1/2)
/
3 (x+1/3)(x-1 ) ≥ 0 ;
методом интервалов :
- + - +
-------------- [ -1/2]//////////// (-1/3) ---------- (1) /////////////////
ответ : x∈ [ -1/2 ; -1/3) ∪ (1 ;∞) .
--------------------------------------------
4.
Вычислить
Log
₅
(
(1/25) ^(-1/
√3) *125^√2
) / (
1/125) ^(-√2)*5 ^(1/√3)
)
=
Log₅
(
5⁻²) ^(-1/√3) *(5³) ^√2
) / (
(5⁻³) ^(-√2)*5 ^(1/√3)
)
=
Log₅ 5^(2/√3)*
5^(3√2)
/ 5^(3√2)
*5 ^(1/√3)
= Log₅ 5 ^(1/√3)
=1/√3 .
1 votes
Thanks 0
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Answers & Comments
Verified answer
:::::::::::::::::::::::::::::::::::::::Verified answer
Task/26450423-------------------
решить неравенство
3.
2 ^(3/(1- x) ) ≤ 0,5 ^(1/(3x+1) )
2 ^(3/(1- x)) ≤ 2 ^(-1/(3x+1) ; т.к. основания 2 > 1 ,то
3 / (1- x) ≤ -1 / (3x+1) ;
3 / (x -1) - 1 / (3x+1) ≥ 0 ;
( 3(3x+1) -(x - 1) ) / (x -1)(3x+1) ≥ 0 ;
8(x +1/2) / 3 (x+1/3)(x-1 ) ≥ 0 ;
методом интервалов :
- + - +
-------------- [ -1/2]//////////// (-1/3) ---------- (1) /////////////////
ответ : x∈ [ -1/2 ; -1/3) ∪ (1 ;∞) .
--------------------------------------------
4.
Вычислить
Log₅ ( (1/25) ^(-1/√3) *125^√2 ) / (1/125) ^(-√2)*5 ^(1/√3) ) =
Log₅ ( 5⁻²) ^(-1/√3) *(5³) ^√2 ) / ( (5⁻³) ^(-√2)*5 ^(1/√3) ) =
Log₅ 5^(2/√3)*5^(3√2) / 5^(3√2)*5 ^(1/√3) = Log₅ 5 ^(1/√3) =1/√3 .