Question

simplifique a expressão acima. ​

Answer 1

Resposta:

Explicação passo-a-passo:

$\frac{A_{6,n}+A_{4,n-2}}{A_{5,n-1}+A_{3,n-3}}=\frac{\frac{6!}{(6-n)!}+\frac{4!}{(4-(n-2))!}}{\frac{5!}{(5-(n-1))}+\frac{3!}{(3-(n-3))}}=\frac{\frac{6!}{(n-6})!+\frac{4!}{(n-6)!}}{\frac{5!}{(n-6)}+\frac{3!}{(n-6)!}}=\frac{\frac{10!}{(n-6})!}{\frac{8!}{(n-6)}}=\frac{10!}{8!}=\frac{10.9.8!}{8!}=10.9=90$