Question

Помогите решить, срочно!

Answer 1

$\sf \dfrac{\dfrac{m-n}{2m-n}-\dfrac{m^2+n^2+m}{2m^2+mn-n^2}}{(4n^4+4mn^2+m^2):(2n^2+m)}=\dfrac{\dfrac{m-n}{2m-n}-\dfrac{m^2+n^2+m}{2m^2+2mn-mn-n^2}}{(2n^2+m)^2:(2n^2+m)}= \\ \\ = \dfrac{\dfrac{m-n}{2m-n}-\dfrac{m^2+n^2+m}{2m(m+n)-n(m+n)}}{2n^2+m}=\dfrac{\dfrac{m-n}{2m-n}-\dfrac{m^2+n^2+m}{(m+n)(2m-n)}}{2n^2+m}= \\ \\ = \dfrac{\dfrac{m^2-n^2-m^2-n^2-m}{(m+n)(2m-n)}}{2n^2+m}=\dfrac{-(2n^2+m)}{(m+n)(2m-n)(2n^2+m)}=\boxed{\sf\dfrac{1}{(n-2m)(m+n)}}$

Answer 2

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