Question

Помогите пожалуйста!
4.51 а

Answer 1

$\\\frac{2mn}{m^3+n^3} + \frac{2m}{m^2-n^2}-\frac{1}{m-n} = \\\frac{2mn}{(m+n)(m^2-mn+n^2)} + \frac{2m}{(m-n)(m+n)}-\frac{1}{m-n} =\\ = \frac{2mn(m-n)+2m(m^2-mn+n^2)-(m+n)(m^2-mn+n^2)}{(m+n)(m-n)(m^2-mn+n^2)} = \\\frac{2m^2n-2mn^2+2m^3-2m^2n+2mn^2-m^3-n^3}{(m+n)(m-n)(m^2-mn+n^2)} = \\\frac{2m^3-m^3-n^3}{(m+n)(m-n)(m^2-mn+n^2)} = \frac{m^3-n^3}{(m+n)(m-n)(m^2-mn+n^2)} =\\=\frac{(m-n)(m^2-mn+n^2)}{(m+n)(m-n)(m^2-mn+n^2)}=\frac{1}{m+n}$