Respuesta :

[tex] \frac{x+3}{x^2-x-12} * \frac{x-4}{x^2-8x+16} \\= \frac{x+3}{(x-4)(x+3)}* \frac{x-4}{(x-4)(x-4)} = \frac{1}{x-4}* \frac{1}{x-4} = \frac{1}{(x-4)^2}[/tex]    *  x-4/x^2-8x+16


Hope that helps