(6m+mn)(-12-2n) = m (6+n) - 2 (6+n) = (6+n)(m-2)
18_03_09_Задание № 1:
Вычислите (2−1)(2+1)(2^2+1)(2^4+1)(2^8+1)−2^16
РЕШЕНИЕ: (2−1)(2+1)(2^2+1)(2^4+1)(2^8+1)−2^16=(2^2−1^2)(2^2+1)(2^4+1)(2^8+1)−2^16=(2^2−1)(2^2+1)(2^4+1)(2^8+1)−2^16=(2^4−1^2)(2^4+1)(2^8+1)−2^16=(2^4−1)(2^4+1)(2^8+1)−2^16=(2^8−1^2)(2^8+1)−2^16=(2^8−1)(2^8+1)−2^16=2^16−1^2−2^16=-1
ОТВЕТ: -1
5)А
=1/2[cos(x+x/2) +cos(x-x/2)]-1/2[cos(x-x/2)-cos(x+x/2]
=1/2[cos(x+x/2) +cos(x-x/2)-cos(x-x/2)+cos(x+x/2)]
=1/2[2cos(x+x/2)]
=cos 3x/2