Question

Simplifique e calcule
A,10!/9!
B,12!/10!
C,99!/101!
D,21!/22!
E,10!/12!
F,4!+5!/4!
G,5!/3!+2!
H,8!-7!/6!

Answer 1

Lambrando algumas regras de fatorial.
$3! = 3*2*1$
$2! = 2*1$
Então $3! = 3*2!$


A. $\frac{10!}{9!} = \frac{10*9!}{9!} = 10$
B. $\frac{12!}{10!} = \frac{12*11*10!}{10!} = 12*11 = 132$
C. $\frac{99!}{101!} = \frac{99!}{101*100*99!} = \frac{1}{101*100} = \frac{1}{10100}$
D. $\frac{21!}{22!} = \frac{21!}{22*21!} = \frac{1}{22}$
E. $\frac{10!}{12!} = \frac{10!}{12*11*10!} = \frac{1}{12*11} = \frac{1}{132}$
F. $\frac{4! + 5!}{4!} = \frac{4! + 5*4!}{4!} = \frac{4!(1 + 5)}{4!} = 1 + 5 = 6$
G. $\frac{5!}{3! + 2!} = \frac{5*4*3*2!}{3*2! + 2!} = \frac{5*4*3*2!}{2!(3 + 1)} = \frac{5*4*3}{3 + 1} = \frac{60}{4} = 15$
H.$\frac{8! - 7!}{6!} = \frac{8*7*6! - 7*6!}{6!} = \frac{56*6! - 7*6!}{6!} = \frac{6!(56 - 7)}{6!} = 56 - 7 = 49$