А второе будет x=5y
итого
x-y=48
x=5y
5y-y=48
4y=48
y=12
x=5*12=60
1) (Cos(1 - x/2))' = -Sin(1 -x/2) * (1 - x/2)' = 1/2*Sin(1 - x/2)
2) (Sin(2 - 3x/4) )' = Cos(2 -3x/4) * (2 -3x/4)'= -3/4*Cos(2 -3x/4)
3) (Sin(x+3)/2) ) ' = Cos(x +3)/2 * ((х +3)/2)' =1/2 * Сos(x +3)/2
4) (Сos(1-x)/3)' = -Sin(1 -x)/3 * ((1-x)/3)' = 1/3* Sin(1-x)/3
5) (Cos(4 -5x)/3)' = -Sin(4-5x)/3 * ( ( 4-5x)/3)' = 5/3 * Sin(4 -5x)/3
6) (Sin(2x +3)/5)' = Cos(2x +3)/5 * ((2x +3)/5)' = 2/5 * Cos(2x+3)/5
7) (Sin²2x)' = 2Sin2x* * (Sin2x)' = 2Sin2x * Cos2x * (2x)' =
=2Sin4x
8) (Cos⁴3x)' = 4Cos³3x * (Cos3x)' = 4Cos³ 3x *(-Sin3x) * (3x)' =
= -12Cos³ 3x* Sin3x
9) (Ctg²4x)' = 2Ctg4x * (Ctg4x)' = 2Ctg4x * (-1/Sin²4x) * (4x)'=
= -8Ctg4x/Sin²4x
10) (tg⁴x/2))' = 4tg³x/2 * (tgx/2)' = 4tg³x/2 * 1/Cos²x/2 * (x/2)'=
=2tg³x/2/Сos²x/2
⁴√(x^2)+5⁴√x-14=0⁴√x=t; t>=0 t^2+5t-14=0D=25+56=81=9^2; t1=(-5-9)/2=-7(посторонний); t2=(-5+9)/2=2
⁴√x=2; x=2^4; x=16Ответ 16
![\displaystyle \frac{1-4\sin^2 \alpha\cos^2\alpha }{(\sin\alpha+\cos\alpha)^2}-2\cos\alpha\sin(-\alpha)= \frac{1-\sin^22\alpha}{1+\sin2\alpha}+2\cos\alpha\sin\alpha=\\ \\ \\ = \frac{(1-\sin2\alpha)(1+\sin2\alpha)}{1+\sin2\alpha}+\sin2\alpha=1-\sin2\alpha+\sin2\alpha=1](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Cfrac%7B1-4%5Csin%5E2+%5Calpha%5Ccos%5E2%5Calpha+%7D%7B%28%5Csin%5Calpha%2B%5Ccos%5Calpha%29%5E2%7D-2%5Ccos%5Calpha%5Csin%28-%5Calpha%29%3D+%5Cfrac%7B1-%5Csin%5E22%5Calpha%7D%7B1%2B%5Csin2%5Calpha%7D%2B2%5Ccos%5Calpha%5Csin%5Calpha%3D%5C%5C+%5C%5C+%5C%5C+%3D+%5Cfrac%7B%281-%5Csin2%5Calpha%29%281%2B%5Csin2%5Calpha%29%7D%7B1%2B%5Csin2%5Calpha%7D%2B%5Csin2%5Calpha%3D1-%5Csin2%5Calpha%2B%5Csin2%5Calpha%3D1+++)
Что и требовалось доказать.