Question

AYUDA PORFAVOR plsssssssssssss​

Answer 1

Respuesta:

$x = 5$

Explicación paso a paso:

$\frac{x - 1}{x - 3} C_{ x - 4}^{x - 2 } + C_{ x - 2}^{x - 1 } + \frac{ x}{x - 1} C_{ x - 2}^{x - 1 } = 15$

$\frac{(x - 1)}{(x - 3)} \frac{(x - 2)! }{(x - 2 - x + 4)! (x - 4)! } + \frac{(x - 1)! }{(x - 1 - x + 2)! (x - 2)! } + \frac{( x)}{(x - 1)} \frac{(x - 1)! }{(x - 1 - x + 2)! (x - 2)!} = 15$

$\frac{(x - 1)}{(x - 3)} \frac{(x - 2)! }{(2)! (x - 4)! } + \frac{(x - 1)! }{(1)! (x - 2)! } + \frac{( x)}{(x - 1)} \frac{(x - 1)! }{(1)! (x - 2)!} = 15$

$\frac{(x - 1)(x - 2)! }{(x - 3)(2)(x - 4)! } + \frac{(x - 1)! }{(x - 2)! } + \frac{(x)(x - 1)! }{(x - 1) (x - 2)!} = 15$

$\frac{(x - 1)(x - 2)(x - 3)(x - 4)! }{(x - 3)(2)(x - 4)! } + \frac{(x - 1)(x - 2)! }{(x - 2)! } + \frac{(x)(x - 1)(x - 2)! }{(x - 1) (x - 2)!} = 15$

$\frac{(x - 1)(x - 2) \cancel{(x - 3)} \cancel{(x - 4)! }}{ \cancel{(x - 3)}(2) \cancel{(x - 4)!} } + \frac{(x - 1) \cancel{(x - 2)! }}{ \cancel{(x - 2)! }} + \frac{(x) \cancel{(x - 1)} \cancel{(x - 2)! }}{ \cancel{(x - 1)} \cancel{ (x - 2)!}} = 15$

$\frac{(x - 1)(x - 2)}{(2) } + (x - 1) +(x) = 15$

$\frac{(x - 1)(x - 2)}{2} + \frac{2(x - 1)}{2} + \frac{2x}{2} = 15$

$(x - 1)(x - 2) + 2(x - 1) + 2x = 15 \cdot 2$

${x}^{2} - 3x + 2+ 2x - 2 + 2x =30$

${x}^{2} + x - 30 = 0$

$(x +6)(x -5)= 0$

$x_1 = -6\: \: , \: \: \: x_2 = 5$

si «x» es positivo

$x = 5$