Решение:
y^3-y^2-y+1=0
(y^3-y^2)-(y-1)=0
y^2(y-1)-(y-1)=0
(y^2-1)(y-1)=0
y^2-1=0
y^2=1
y1,2=+-√1=+-1
y1=1
y2=-1
y-1=0
y=1
Ответ: у1=1; у2=-1
(13-√13)√13/13=13√13-13/13=13(√13-1)/13=√13-1
А)х=3 х=-5
б)х=0 х=1/3 х=-1/2
cos(α)+sin(α)-cos(2π);
1)cos(α)-cos(2π)=-2sin((α+2π)/2)sin((α-2π)/2)=-2sin(α/2+π)sin(α/2-π)=
=2sin(π+α/2)sin(π-α/2)=-2sin(α/2)sin(α/2);
2)sin(α)=2sin(α/2)cos(α/2);
1+2) 2sin(α/2)cos(α/2)-2sin(α/2)sin(α/2)=2sin(α/2)(cos(α/2)-sin(α/2));