Question

6 орнекти ыкшамданыз.

Answer 1

Ответ:

$\Big(\dfrac{mn+n^2}{4m^2-4mn}+mn+n^2\Big)\cdot \dfrac{4m}{m+n}-\dfrac{n}{m-n}=\\\\\\=\Big(\dfrac{n(m+n)}{4m(m-n)}+n(m+n)\Big)\cdot \dfrac{4m}{m+n}-\dfrac{n}{m-n}=$

$=n(m+n)\cdot \Big(\dfrac{1}{4m(m-n)}+1\Big)\cdot \dfrac{4m}{m+n}-\dfrac{n}{m-n}=\\\\\\=n(m+n)\cdot \dfrac{1+4m(m-n)}{4m(m-n)}\cdot \dfrac{4m}{m+n}-\dfrac{n}{m-n}=\\\\\\=\dfrac{n\cdot (1+4m^2-4mn)}{m-n}-\dfrac{n}{m-n}=\dfrac{n+4nm^2-4mn^2-n}{m-n}=\\\\\\=\dfrac{4mn(m-n)}{m-n}=4mn$