me pueden ayudar a despejar "x"
(x-1)! + 2(x+1)!
-------------------- = 13
x! - (x-1)!
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Factor común (x-1)! + 2(x+1)x! = 13
x(x-1)! - (x-1)!
(x-1)! + 2(x+1)(x)(x-1)! = 13
x(x-1)! - (x-1)!
(x-1)! [1+ 2(x+1)(x)] = 13
x(x-1)! - (x-1)!
Simplifico (x-1)!
1+ 2(x^2+x) = 13
x-1
2x^2+2x+1 = 13
x-1
Tenemos ahora: 2x^2+2x+1 = 13(x-1)
2x^2+2x+1 = 13x -13
2x^2 - 11x+14 = 0
(2x - 7 )(2x - 4)
-------------------- = 0
2
(2x-7)(x-2)=0
x = 7/2 descartado porque factorial es para enteros positivos
x= 2
x(x-1)! - (x-1)!
(x-1)! + 2(x+1)(x)(x-1)! = 13
x(x-1)! - (x-1)!
(x-1)! [1+ 2(x+1)(x)] = 13
x(x-1)! - (x-1)!
Simplifico (x-1)!
1+ 2(x^2+x) = 13
x-1
2x^2+2x+1 = 13
x-1
Tenemos ahora: 2x^2+2x+1 = 13(x-1)
2x^2+2x+1 = 13x -13
2x^2 - 11x+14 = 0
(2x - 7 )(2x - 4)
-------------------- = 0
2
(2x-7)(x-2)=0
x = 7/2 descartado porque factorial es para enteros positivos
x= 2
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