Упростите выражение: 4) (b^2+2)/(b-5) + 27/(5-b)

а) A(n) = 2 - 3 · n;  

A(1) = -1

A(2) = -4

A(3) = -7

A(4) = -10

A(5) = -13

б) A(n) = 50 - 7 · n;  

A(1) = 43

A(2) = 36

A(3) = 29  

A(4) = 22

A(5) = 15

в) B(n) = 1 ÷ n + 1;

B(1) = 2

B(2) = 1,5

B(3) =  

B(4) = 1,25

B(5) = 1,2

г) B(n) = n³

B(1) = 1

B(2) = 8

B(3) = 27

B(4) = 64

B(5) = 125

Объяснение:

а) A(n) = 2 - 3 · n;  

A(1) = 2 - 3 · 1 = -1

A(2) = 2 - 3 · 2 = -4

A(3) = 2 - 3 · 3 = -7

A(4) = 2 - 3 · 4 = -10

A(5) = 2 - 3 · 5 = -13

б) A(n) = 50 - 7 · n;  

A(1) = 50 - 7 · 1 = 43

A(2) = 50 - 7 · 2 = 36

A(3) = 50 - 7 · 3 = 29  

A(4) = 50 - 7 · 4 = 22

A(5) = 50 - 7 · 5 = 15

в) B(n) = 1 ÷ n + 1;

B(1) = 1 ÷ 1 + 1 = 2

B(2) = 1 ÷ 2 + 1 = 1,5

B(3) = 1 ÷ 3 + 1 =

B(4) = 1 ÷ 4 + 1 = 1,25

B(5) = 1 ÷ 5 + 1 = 1,2

г) B(n) = n³

B(1) = 1³ = 1

B(2) = 2³ = 8

B(3) = 3³ = 27

B(4) = 4³ = 64

B(5) = 5³ = 125

1
(cos²2t-sin²2t)(cos²2t+sin²2t)=cos²2t-sin²2t=cos4t
2
sina=-√(1-cos²a)=-√(1-225/289)=-√(64/289)=-8/17
sin2a=2sinacosa=2*(-8/17)*15/17=-240/289
cos2a=cos²a-sin²a=225/289-64/289=161/289
tg2a=sin2a/cos2a=-240/289:161/289=-240/289*289/161=-240/161
3
(sin3acos2b+cos3asin2b-sin3acos2b+cos3asin2b)/(cos3acos2b-sin3asin2b+
+cos3acos2b+sin3asin2b)=2cos3asin2b/2cos3acos2b=sin2b/cos2b=tg2b
4
sina=-√(1-cos²a)=-√(1-4/9)=-√5/3
cosb=-√(1-sin²b)=-√(1-1/9)=-2√2/3
sin2a=2sinacosa=2*(-√5/3)*2/3=-4√5/9
sin2b=2sinbcosb=2*1/3*(-2√2/3)=-4√2/9
cos2a=cos²a-sin²a=4/9-5/9=-1/9
cos2b=cos²b-sin²b=8/9-1/9=7/9
sin(2a+2b)=sin2acos2b+cos2asin2b=-4√5/9*7/9-1/9*(-4√2/9)=
=-28√5/81+4√2/81=4(√2-7√5)/81