Question

Fatorial! Calcule: a) 7! / 4!+3!
Simplifique: b) (n+4)! / (n+2)!+(n+3)!
c) n!-(n-2)! / (n-3)!
d) (n-1)! / n! * (n+2)! / (n-3)!
e) (n-1)! +n! / (n+1)!

Answer 1

a) 
$\frac{7!}{4!+3!} = \frac{7*5*4!}{4!+3!} = \frac{7*5}{3!}= \frac{7*5}{3*2*1}= \frac{35}{6}$
b)
$\frac{(n+4)!}{(n+2)!+(n+3)!} = \frac{(n+4)*(n+3)!}{(n+2)!+(n+3)!} =\frac{(n+4)}{(n+2)!} =\frac{(n+4)}{(n+2)*(n+1)} =\frac{(n+4)}{(n^2+n+2n+2)}=$

$=\frac{(n+4)}{(n^2+3n+2)}==\frac{n+4}{n^2+3n+2}=$

Vou fazer só essas duas as outras é o mesmo processo da letra b

Answer 2

Resposta:

a) 7!/4!+3 = 7x6x5x4/4!3! = 7x6x5/3x2 = 210/6 =35

Explicação passo-a-passo: