Ответ: \[\frac{4}{m-8}\] доказано
\[\left(\frac{m}{m^2-16m+64} - \frac{m+4}{m^2-64}\right) : \frac{3m+8}{m^2-64} = \frac{4}{m-8}\]
\[\left(\frac{m}{(m-8)^2} - \frac{m+4}{(m-8)(m+8)}\right) : \frac{3m+8}{(m-8)(m+8)}\]
\[\left(\frac{m(m+8)}{(m-8)^2(m+8)} - \frac{(m+4)(m-8)}{(m-8)^2(m+8)}\right) : \frac{3m+8}{(m-8)(m+8)}\]
\[\left(\frac{m^2+8m - (m^2-8m+4m-32)}{(m-8)^2(m+8)}\right) : \frac{3m+8}{(m-8)(m+8)} = \frac{m^2+8m - m^2+4m+32}{(m-8)^2(m+8)} : \frac{3m+8}{(m-8)(m+8)} = \frac{12m+32}{(m-8)^2(m+8)} : \frac{3m+8}{(m-8)(m+8)}\]
\[\frac{4(3m+8)}{(m-8)^2(m+8)} \cdot \frac{(m-8)(m+8)}{3m+8}\]
\[\frac{4(3m+8)(m-8)(m+8)}{(m-8)^2(m+8)(3m+8)} = \frac{4}{m-8}\]
\[\frac{4}{m-8} = \frac{4}{m-8}\]
Ответ: \[\frac{4}{m-8}\] доказано