Question

упростить (n+2)!(n^2-9)/(n+4)!

Answer 1

(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)