Question

(x-4)!(x+4)! sobre (x-5)!(x+3)! = 9​

Answer 1

Antes de resolver los factoriales, sabemos que:

(x+a)! = (x+a)(x+a-1)!

Entonces:

$\frac{(x-4)!(x+4)!}{(x-5)!(x+3)!} = 9\\\\ \frac{(x-4)(x-4-1)!(x+4)(x+4-1)!}{(x-5)!(x+3)!} = 9\\\\\\ \frac{(x-4)(x-5)!(x+4)(x+3)!}{(x-5)!(x+3)!} = 9\\\\ \ (x-4)(x+4)= 9\\\\ x^{2}-16 = 9\\x^{2} = 9 +16\\x^{2} = 25\\x = 5$