Question

$\frac{5}{1 + \frac{4}{x} } * ( \frac{x-4}{x^2+4x} - \frac{16}{16-x^2} )$

Answer 1

$\frac{5}{1+\frac{4}{x}}*(\frac{x-4}{x^2+4x}-\frac{16}{16-x^2})=\frac{5x}{x+4}*\frac{(x-4)^2-16x}{x(x+4)(x-4)}=\frac{5x(x^2-24x+16)}{x(x+4)^2(x-4)}=\frac{5(x^2-24x+16)}{(x+4)^2(x-4)}$