Вопрос от посетителя
упростить (n+2)!(n^2-9)/(n+4)!
Отвечает Илюха:
(n+2)!(n^2-9)/(n+4)!= (n+2)!(n^2-9)/((n+2)!(n+3)(n+4))=(n^2-9)/((n+3)(n+4))=
=(n^2-9)/(n^2+7n+12)=(n^2+7n+12-7n-21)/(n^2+7n+12)=1 – (7n+21)/(n^2+7n+12)
упростить (n+2)!(n^2-9)/(n+4)!
(n+2)!(n^2-9)/(n+4)!= (n+2)!(n^2-9)/((n+2)!(n+3)(n+4))=(n^2-9)/((n+3)(n+4))=
=(n^2-9)/(n^2+7n+12)=(n^2+7n+12-7n-21)/(n^2+7n+12)=1 – (7n+21)/(n^2+7n+12)