Question

Помогите решить, срочно!

Answer 1

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

Answer 2

Спасибо огромное!

Answer 3

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