Question

Em cada caso, calcule o valor de:
a)
$\frac{4!}{4! + 8!}$
b)
$\frac{8!}{4! + 8!}$

Answer 1

Olá, tudo bem?

a)

$\dfrac{4!}{4!+8!}=\dfrac{\not4!}{\not4![1+(8\cdot 7\cdot 6\cdot 5)]}=\dfrac{1}{1681}$

b)

$\dfrac{8!}{4!+8!}=\dfrac{4!\cdot8\cdot 7\cdot 6\cdot 5}{4!+4!\cdot 8\cdot 7\cdot 6\cdot 5}=\\\\\\\dfrac{\not 4!\cdot8\cdot 7\cdot 6\cdot 5}{\not 4![1+(8\cdot 7\cdot 6\cdot 5)]}=\dfrac{1680}{1681}$

É isso!! :)