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(x+1) (x!) (x-1)! = (x!)² + 36x
(x+1)(x)(x-1)!(x-1)! = (x!) (x!) +36x
(x)(x+1)((x-1)!)² = x(x-1)! (x)(x-1)! +36x
(x)(x+1)((x-1)!)² = (x) (x((x-1)!)² +36)
(x+1)((x-1)!)² = (x)((x-1)!)² + 36
n = (x-1)!
(x+1)n² = xn² +36
(x+1)n² - xn² - 36 = 0
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(x +1)-(x)
= x + 1 - x
= 1
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n² - 36 = 0
(n - 6) (n + 6) = 0
n = ±6
Al reemplazar tomamos solo el valor positivo puesto que el factorial no puede ser un número negativo.
Entonces:
(x-1)! = 6
Sabemos que 3! = 6 por ende
(x-1) = 3
x = 3 + 1
x = 4
(4-1)! = 3! = 6
Comprobando:
"x = 4"
(x+1)! (x-1)! = 36x + (x!)²
(4+1)! (4-1)! = 36(4) + (4!)²
5! * 3! = 144 + (4!)²
(5*4*3*2*1) (3*2*1) = 144 + (4*3*2*1)²
(120) (6) = 144 + (24)²
720 = 144 + 576
720 = 720