Ответ:

[tex]\frac{m}{m-3}[/tex]

Пошаговое объяснение:

[tex]\frac{2m}{m-3}+\frac{m}{m+3}+\frac{2m^2}{9-m^2}=\\\\=\frac{2m}{m-3}+\frac{m}{m+3}-\frac{2m^2}{m^2-9}=\\\\=\frac{2m}{m-3}+\frac{m}{m+3}-\frac{2m^2}{(m-3)(m+3)}=\\\\=\frac{2m(m+3)+m(m-3)-2m^2}{(m-3)(m+3)}\\\\=\frac{2m^2+6m+m^2-3m-2m^2}{(m-3)(m+3)}=\\\\=\frac{m^2+3m}{(m-3)(m+3)}=\\\\=\frac{m(m+3)}{(m-3)(m+3)}=\\\\=\frac{m}{m-3}[/tex]