Объяснение: * * * cos(-α) =cosα , sin2α=2sinα*cosα и формулы
приведения * * *
1) ( 1 +sin(4π -(π/2 +α) )+ cos(2π-2α) ) / (2sinα*cosα - sinα) =
( 1 - cosα+ cos2α ) / (2sinα*cosα - sinα) =(2cos²α -cosα) / (2cosα -1)sinα=
(2cosα -1)cosα / (2cosα -1)sinα = ctgα .
2) ( sin²α - 4sin²(α/2) ) / ( sin²α - 4+4sin²(α/2) ) =
( 4sin²(α/2) cos²(α/2) - 4sin²(α/2) ) / ( 4sin²(α/2) cos²(α/2) - 4(1 -sin²(α/2) ) =
- 4sin²(α/2) (1 - cos²(α/2) ) / - 4cos²(α/2)( 1 -sin²(α/2) ) =
sin⁴(α/2) / cos⁴(α/2)= tg⁴(α/2) .
3) (cos²α -sin²α ) / (1+sin2α) =
|| * * * 1+sin2α= cos²α+sin²α+2sinαcosα =(cosα+sinα)² * * * ||
= (cosα -sinα )(cosα+sinα) /(cosα+sinα)²= (cosα -sinα)/(cosα+sinα)
4) ( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
|| sinα -sinβ =2sin( (α -β)/2 ) *cos( (α +β)/2 ) ||
|| sinα+sinβ =2sin( (α+β)/2 ) *cos( (α -β)/2 ) ||
= 2 sinβcosα / 2 sinαcosβ =(cosα / sinα) *(sinβ/cosβ) = ctgα *tgβ =
ctgα / ctg β.
* * * * * * * по другому * * * * * * *
( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
( sinαcosβ+cosα*sinβ - (sinαcosβ-cosα*sinβ) ) *
1/ ( sinαcosβ+cosα*sinβ+ sinαcosβ-cosα*sinβ ) =
2cosα*sinβ /2 sinαcosβ =ctgα /ctgβ
* * * * * * * * * * * * * * * * * * * * *
5) ( (1+cosx)/sinx )*(1+ ( (1 -cosx)/sinx )² ) =
( (1+cosx)/sinx )*(sin²x +1 -2cosx+cos²x )/sin²x ) =
( (1+cosx)/sinx )*( 2(1 -cosx))/sin²x ) = 2(1+cosx)(1-cosx) /sin³x =
2(1 - cos²x) /sin³x =2sin²x/ sin³x = 2 / sinx .
= ( 2cos²(x/2) / 2sin(x/2)*cos(x/2) )*(1+ ( 2sin²(x/2) / 2sin(x/2)*cos(x/2) )² ) =
(cos(x/2) / sin(x/2) )*( 1 + sin²(x/2) / cos²(x/2) ) =
(cos(x/2) /sin(x/2) )*( ( cos²(x/2) + sin²(x/2) ) /cos²(x/2) ) =
(cos(x/2) /sin(x/2) )* ( 1 / cos²(x/2) ) = 1 /( cos(x/2)*sin(x/2) ) =2/sinx
1) n=7
2) n=4
3) k=87
4) x=3
Объяснение:
Формула n!=1·2·3·...·n, исключение 0!=1, n!>0, n=0;1;2;3;...
1) n!=7(n-1)!
n·(n-1)!=7(n-1)!
Делим на (n-1)!>0
n=7
2) (n+17)!=420(n+15)!
(n+15)!(n+16)(n+17)=420(n+15)!
Делим на (n+15)!>0
(n+16)(n+17)=420
n²+33n+272=420
n²+33n-148=0
D=33²-4·1·(-148)=1089+592=1681=41²
n₁,₂=(-33±41)/2
n₁=(-33-41)/2=-37<0
n₂=(-33+41)/2=4
3) (k-10)!=77(k-11)!
(k-10)(k-11)!=77(k-11)!
k-10=77
k=87
4) (3x)!=504(3x-3)!
x≥3
3x(3x-1)(3x-2)(3x-3)!=504(3x-3)!
Делим на 3(3x-3)!>0
x(3x-1)(3x-2)=168
9x³-9x²+2x-168=0
9x³-27x²+18x²-54x+56x-168=0
9x²(x-3)+18x(x-3)+56(x-3)=0
(x-3)(9x²+18x+56)=0
x-3=0
x=3
9x²+18x+56=0
D=324-224=100=10²
x₁=(-18-10)/18=-14/9<0
x₂=(-18+10)/18=-4/9<0
Объяснение: * * * cos(-α) =cosα , sin2α=2sinα*cosα и формулы
приведения * * *
1) ( 1 +sin(4π -(π/2 +α) )+ cos(2π-2α) ) / (2sinα*cosα - sinα) =
( 1 - cosα+ cos2α ) / (2sinα*cosα - sinα) =(2cos²α -cosα) / (2cosα -1)sinα=
(2cosα -1)cosα / (2cosα -1)sinα = ctgα .
2) ( sin²α - 4sin²(α/2) ) / ( sin²α - 4+4sin²(α/2) ) =
( 4sin²(α/2) cos²(α/2) - 4sin²(α/2) ) / ( 4sin²(α/2) cos²(α/2) - 4(1 -sin²(α/2) ) =
- 4sin²(α/2) (1 - cos²(α/2) ) / - 4cos²(α/2)( 1 -sin²(α/2) ) =
sin⁴(α/2) / cos⁴(α/2)= tg⁴(α/2) .
3) (cos²α -sin²α ) / (1+sin2α) =
|| * * * 1+sin2α= cos²α+sin²α+2sinαcosα =(cosα+sinα)² * * * ||
= (cosα -sinα )(cosα+sinα) /(cosα+sinα)²= (cosα -sinα)/(cosα+sinα)
4) ( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
|| sinα -sinβ =2sin( (α -β)/2 ) *cos( (α +β)/2 ) ||
|| sinα+sinβ =2sin( (α+β)/2 ) *cos( (α -β)/2 ) ||
= 2 sinβcosα / 2 sinαcosβ =(cosα / sinα) *(sinβ/cosβ) = ctgα *tgβ =
ctgα / ctg β.
* * * * * * * по другому * * * * * * *
( sin(α+β) - sin(α -β) ) / ( sin(α+β) +sin(α -β) ) =
( sinαcosβ+cosα*sinβ - (sinαcosβ-cosα*sinβ) ) *
1/ ( sinαcosβ+cosα*sinβ+ sinαcosβ-cosα*sinβ ) =
2cosα*sinβ /2 sinαcosβ =ctgα /ctgβ
* * * * * * * * * * * * * * * * * * * * *
5) ( (1+cosx)/sinx )*(1+ ( (1 -cosx)/sinx )² ) =
( (1+cosx)/sinx )*(sin²x +1 -2cosx+cos²x )/sin²x ) =
( (1+cosx)/sinx )*( 2(1 -cosx))/sin²x ) = 2(1+cosx)(1-cosx) /sin³x =
2(1 - cos²x) /sin³x =2sin²x/ sin³x = 2 / sinx .
* * * * * * * по другому * * * * * * *
= ( 2cos²(x/2) / 2sin(x/2)*cos(x/2) )*(1+ ( 2sin²(x/2) / 2sin(x/2)*cos(x/2) )² ) =
(cos(x/2) / sin(x/2) )*( 1 + sin²(x/2) / cos²(x/2) ) =
(cos(x/2) /sin(x/2) )*( ( cos²(x/2) + sin²(x/2) ) /cos²(x/2) ) =
(cos(x/2) /sin(x/2) )* ( 1 / cos²(x/2) ) = 1 /( cos(x/2)*sin(x/2) ) =2/sinx
* * * * * * * * * * * * * * * * * * * * *
1) n=7
2) n=4
3) k=87
4) x=3
Объяснение:
Формула n!=1·2·3·...·n, исключение 0!=1, n!>0, n=0;1;2;3;...
1) n!=7(n-1)!
n·(n-1)!=7(n-1)!
Делим на (n-1)!>0
n=7
2) (n+17)!=420(n+15)!
(n+15)!(n+16)(n+17)=420(n+15)!
Делим на (n+15)!>0
(n+16)(n+17)=420
n²+33n+272=420
n²+33n-148=0
D=33²-4·1·(-148)=1089+592=1681=41²
n₁,₂=(-33±41)/2
n₁=(-33-41)/2=-37<0
n₂=(-33+41)/2=4
3) (k-10)!=77(k-11)!
(k-10)(k-11)!=77(k-11)!
k-10=77
k=87
4) (3x)!=504(3x-3)!
x≥3
3x(3x-1)(3x-2)(3x-3)!=504(3x-3)!
Делим на 3(3x-3)!>0
x(3x-1)(3x-2)=168
9x³-9x²+2x-168=0
9x³-27x²+18x²-54x+56x-168=0
9x²(x-3)+18x(x-3)+56(x-3)=0
(x-3)(9x²+18x+56)=0
x-3=0
x=3
9x²+18x+56=0
D=324-224=100=10²
x₁=(-18-10)/18=-14/9<0
x₂=(-18+10)/18=-4/9<0