Ответ:
10,125.
Объяснение:
27(-3^2а^3):(3^5а^-1)^3 при а=-0,1
27(-18а):(-16a)3
-486а:(-48а)
При а=(-0,1): -486(-0,1):48^0,1
При а=(-0,1): 48,6:4,8
При а=(-0,1): 10,125.
Ответ: 10,125
1) т.к. 1.5π<α<2π
то cosα>0
⇒cosα = √(1-sin²α) = √(1-9/25) = 4/5
2)sin(π-α)=sinα
sin(π-α)=sinπ*cosα - sinα*cosπ = [т.к. sinπ=0 и cosπ=-1] = sinπ
3) sin(11π/4) = sin(3π/4) = √2/2
cos(13π/4) = cos(π/4) = √2/2
sin(-2.5π) = sin(-0.5π) = sin(-π/2) = -1
cos(-25π/3) = cos(25π/3) = cos(π/3) = 1/2
(√2/2 - √2/2) *(-1) * (1/2) = 0
4) cosα=-2/3
sinα = ±√(1-cos²α) = ±√(1-4/9) = ±√5/3
⇒|sinα|<1
√((1-sinα)/(1+sinα))=√((1-sinα)²/(1-sin²α))=√((1-sinα)²/cos²α)=|(1-sinα)|/|cosα|
√((1+sinα)/(1-sinα))=√((1+sinα)²/(1-sin²α))=√((1+sinα)²/cos²α)=
=|(1+sinα)|/|cosα|
|(1-sinα)|/|cosα| + |(1+sinα)|/|cosα| = (|1-sinα|+|1+sinα|)/|cosα| =
=(1-sinα+1+sinα)/|cosα| = 2/|cosα| = 2/ (2/3) = 3
1) x = 5
2) x = 1/3
3) x = 125
4) x = 8
5) x принадлежит промежутку (2; +бесконечности)
6) x принадлежит промежутку (0; 1)
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