$$\frac{(x-4)^2}{16-x^2}=\frac{(x-4)^2}{(4-x)(4+x)}=\frac{-(4-x)^2}{(4-x)(4+x)}=-\frac{4-x}{4+x}$$
Ответ: $$\frac{-(4-x)}{4+x}$$