Answer :

[tex]\frac{\frac{x+4}{x^2-16}}{\frac{x^2-1}{x^2+3x-28}}\\=\frac{x+4}{(x+4)(x-4)}\times\frac{(x-4)(x+7)}{(x-1)(x+1)}\\=\frac{x+7}{x^2-1}[/tex]

Other Questions