решение во вложениях)))))))))))))
<span>(2x+y)*(4x^2-2xy+y^2)-y^2*(y-1)-7x^3=</span>8x^3+y^3-y^3+y^2-7x^3=8x^3+y^2-7x^3=x^3+y^2
1) sin 390 =sin(360+30)=sin30= 1/2
2) cos420=cos(360+60)=cos60=1/2
3) tg 540=tg(360+180)=tg180=0
4) ctg450=ctg(360+90)=ctg90=0
5) tg 7π/3 = tg(2π + π/3)=tg π/3 =√3
6) sin 11π/6 = sin(2π - π/6) = - sin π/6 = - 1/2
7) cos 9π/4 = cos(2π + π/4) = cos π/4 = √2/2
8) ctg 10π/3 = ctg (4π - 2π/3)= - ctg 2π/3 = -ctg(π - π/3) =ctg π/3 = √3/3
У'=5х^4+((х^5-1)-5х^5)/(х^5-1)^2=-(1+6х^5)/(х^5-1)^2
Вроде так