Помогите решить, срочно!

Решите задачу:

$\frac{\frac{m-n}{2m-n}-\frac{m^2+n^2+m}{2m^2+mn-n^2} }{\frac{4n^4+4mn^2+m^2}{2n^2+m} }= \frac{\frac{m-n}{2m-n}-\frac{m^2+n^2+m}{2m^2+2mn-mn-n^2} }{\frac{4n^4+4mn^2+m^2}{2n^2+m} }=\frac{\frac{m-n}{2m-n}-\frac{m^2+n^2+m}{2m*(m+n)-n(m+n)} }{\frac{4n^4+4mn^2+m^2}{2n^2+m} }=\\=\frac{\frac{m-n}{2m-n}-\frac{m^2+n^2+m}{(2m-n)(m+n)}*(2n^2+m) }{{4n^4+4mn^2+m^2} }=\frac{\frac{(m+n)(m-n)-m^2-n^2-m}{(2m-n)(m+n)} *(2n^2+m)}{4n^4+4mn^2+m^2}=\frac{(-2n^2-m)(2n^2+m)}{(2m-n)(m+n)(2n^2+m)^2} =\\=-\frac{1}{(2m-n)(m+n)}$