Помогите пожалуйста не могу решить

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

$\frac{r^2-25}{r+3} *\frac{1}{r^2+5r} -\frac{r+5}{r^2-3r} =\frac{(r-5)(r+5)}{r(r+3)(r+5)} -\frac{r+5}{r(r-3)} =\frac{(r-5)(r-3)-(r+5)(r+3)}{r(r+3)(r-3)} =\\\frac{r^2-8r+15-r^2-8r-15}{r(r-3)(r+3)} =-\frac{16}{(r-3)(r+3)}$

$(\frac{st}{s^2-t^2} +\frac{t}{2t-2s}) \frac{s^2-t^2}{2t}= (\frac{st}{(s-t)(s+t)}- \frac{t}{2(s-t)} ) \frac{s^2-t^2}{2t}=\\\frac{2st-st-t^2}{2(s-t)(s+t)}* \frac{s^2-t^2}{2t}=\frac{t(s-t)}{2(s-t)(s+t)} *\frac{(s-t)(s+t)}{2t} =\frac{s-t}{4}$

$\frac{a+b}{3a+b} +\frac{1}{b-a} *\frac{a^2-b^2}{3a-b}= \frac{a+b}{3a+b}- \frac{(a-b)(a+b)}{(a-b)(3a-b)} =\frac{a+b}{3a+b} -\frac{a+b}{3a-b} =\\\frac{(a+b)(3a-b-3a-b)}{(3a+b)(3a-b)} =\frac{-2b(a+b)}{(3a-b)(3a+b)}$