Сократите дробь a) (n+1)!/n! b) n!/(n+2)! B) (n+1)! (n+3)/(n+4)!
a) (n+1)!/n!=(n+1)n!/n!=(n+1)
b) n!/(n+2)!=n!/((n+2)n!)=1/(n+2)
B) (n+1)!(n+3)/(n+4)!= (n+1)!(n+3)/((n+1)!(n+2)(n+3)(n+4))=1/((n+2)(n+4))