X^3-y^3+3y^2-3y+1=x^3-(y^3-3y^2+3y-1)=x^3-(y-1)^3=(x-y+1)(x^2+x(y-1)+(y-1)^2)=(x-y+1)(x^2+xy-x+y^2-2y+1)<span>
</span>8x^3+y^3+6y^2+12y+8=(2x)^3+(y+2)^3=(2x+y+2)(4x^2-2x(y+2)+(y+2)^2)=(2x+y+2)(4x^2-2xy-4x+y^2+4y+4)=<span>(2x+y+2)(4x^2-2xy-4x+y^2+4y+4)</span>
3^5*3^6 / 3^4*3^2=3^5*3^6 / 3^6=3^5. Ответ: 3^5.
Замена: (x^2+2x)=t
получаем:
t(t-2)=3
t^2 -2t-3=0
t1=3 t2=-1
1) x^2+2x=3
x1=-3 x2=1
2) x^2+2x=-1
x1=-1 x2=-1
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