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>
3/5pk^3*(10p-5p^3k-1/3k^2) == 6/k^3 - 3p^2k^4 - 5pk^5
1/4с^2d^2*(4c^2-2cd^2+d) ==c^4d^2 -1/2c^3d^3 + 1/4c^2d^3
(5 - с)² = 5² - 10c + c² = 25 - 10 c + c²
P.S. Формула: (a - b)² = a² - 2ab + b²