Решение
2cos ( 2x + пи/3) <= 1
cos ( 2x + пи/3) <= 1/2
пи/3 +2пи*n <= 2x + пи/3<= 5пи/3+2пи*n
2пи*n <= 2x <= 4пи/3+2пи*n
<span>пи*n <= x <= 2пи/3+пи*n</span>
В)
x⁴+y⁴=
(x+y)⁴=(x²+2xy+y²)=x⁴+4x³y+4xy³+6x²y²+y⁴=(x⁴+y⁴)+6x²y²+4xy(x²+y²)=(x⁴+y⁴)+6x²y²+4xy((x+y)²-2xy) ⇒
x⁴+y⁴=(x+y)⁴-(6x²y²+4xy((x+y)²-2xy))=(-3)⁴-(6*(-5)²+4(-5)((-3)²-2(-5))=81-20(9+10)=81-20*19=-299
г)
x²y²(x⁵+y⁵)=(-5)²(x⁵+y⁵)=25(x⁵+y⁵)=
(x+y)⁵=x⁵+5x⁴y+10x²y³+10x³y²+5xy⁴+y⁵=(x⁵+y⁵)+5xy(x³+y³)+10x²y²(x+y)=
(x⁵+y⁵)+5xy(x+y)(x²+y²-xy)+10x²y²(x+y)=(x⁵+y⁵)+5xy(x+y)((x+y)²-3xy)+10x²y²(x+y) ⇒
x⁵+y⁵=(x+y)⁵-(5xy(x+y)((x+y)²-3xy)+10x²y²(x+y))=(-3)⁵-(5*(-5)(-3)((-3)²+15)+10*(-5)²(-3))=-243-(1800-750)=-243-1050=-1293