Question

simplifique

a) (n+3)! -2(n+3)!
___________
(n+2)!-n!


b) 2(n+1)!+3(n-1)!
___________
(n+2)!

peço ajuda Pfv
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Answer 1

Resposta:

A)

$\frac{(n+3)!-2(n+3)!}{(n+2)!-n!}$$=$

$=\frac{(n+3)*(n+2)*(n+1)*n!-2*(n+3)*(n+2)*(n+1)*n!}{(n+2)*(n+1)*n!-n!} =$

$=\frac{n![(n+3)*(n+2)*(n+1)-2*(n+3)*(n+2)*(n+1)]}{n![(n+2)*(n+1)-1]}=$

$=\frac{(n+3)*(n+2)*(n+1)[1-2]}{n^2+n+2n+2-1}=$

$=\frac{-[(n+3)*(n+2)*(n+1)]}{n^2+3n+1} =$

$=\frac{-[(n+3)*(n^2+3n+2)]}{n^2+3n+1}=$

$=\frac{-(n^3+3n^2+2n+3n^2+9n+6)}{n^2+3n+1}=$

$=\frac{-n^3-6n^2-11n-6}{n^2+3n+1}$

B)

$\frac{2(n+1)!+3(n-1)!}{(n+2)!}=$

$=\frac{2*(n+1)!+3*(n+1)!}{(n+2)*(n+1)!}=$

$=\frac{2+3}{n+2}=$

$=\frac{5}{n+2}$

Espero ter ajudado!!

Qualquer duvida, comente.