Запишите в виде дроби а)n+2/n! -3n+2/(n+1)! б)1/(k-/(k+1) в)1/(k-+k/(k+1)!

А
(n+2)/n!-(3n+2)/(n+1)! =(n+2)/n!-(3n+2)/[n!(n+1)]=
=(n²+n+2n+2-3n-2)/[n!(n+1)]=n²/(n+1)!
Б
1/(k-1)!-k/(k+1)!=1/(k-1)!-k/[(k-1)!*k*(n+1)]=
=(k²+k-k)/(k+1)!=k²/(k+1)!
В
1/(k-2)!-(k^3+k)/(k+1)! =1/(k-2)!-(k³+k)/[(k-2)!(k-1)k(k+1)]=
=(k³-k-k³-k)/(k+1)!=-2k/(k+1)!