Paso a paso:
(x + 2)! = (x + 2) (x + 1)!
(x + 3)! = (x + 3) (x + 2) (x + 1)!
De la expresión del problema se cancela (x + 1)! quedando:
[(x + 2) + (x + 3) (x + 2)] / [(x + 2) - 1] = 64 / (x + 1)
Quitamos corchetes:
(x² + 6 x + 9) / (x + 1) = 64 / (x + 1); simplificamos:
(x + 3)² = 64; entonces:
x + 3 = ± 8
x + 3 = 8; resulta x = 5
x + 3) = - 8; resulta x = - 11
No existen factoriales de números negativos.
Verificamos:
(5 + 1)! = 720; (5 + 2)! = 5040; (5 + 3)! = 40320
(720 + 5040 + 40320) / (5040 - 720) = 46080 / 4320 = 32/3 = 64/6
Mateo
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Verified answer
Paso a paso:
(x + 2)! = (x + 2) (x + 1)!
(x + 3)! = (x + 3) (x + 2) (x + 1)!
De la expresión del problema se cancela (x + 1)! quedando:
[(x + 2) + (x + 3) (x + 2)] / [(x + 2) - 1] = 64 / (x + 1)
Quitamos corchetes:
(x² + 6 x + 9) / (x + 1) = 64 / (x + 1); simplificamos:
(x + 3)² = 64; entonces:
x + 3 = ± 8
x + 3 = 8; resulta x = 5
x + 3) = - 8; resulta x = - 11
No existen factoriales de números negativos.
Respuesta: x = 5
Verificamos:
(5 + 1)! = 720; (5 + 2)! = 5040; (5 + 3)! = 40320
(720 + 5040 + 40320) / (5040 - 720) = 46080 / 4320 = 32/3 = 64/6
Mateo