Respuesta :

[tex] \frac{x^2+4x+3}{x^2-4x-21} = \frac{(x+3)(x+1)}{(x+3)(x-7)} = \frac{x+1}{x-7} [/tex]
[tex] \frac{x^2+10x+16}{x^2+13x+40} = \frac{(x+8)(x+2)}{(x+8)(x+5)} = \frac{x+2}{x+5} [/tex]