ПОМОГИТЕ, В ТРЕТИЙ РАЗ ПРОШУ,НОМЕР 8

8.
1)
$\frac{2m}{(2m-3)^2}+ \frac{3-m}{9-4m^2}= \frac{2m}{(2m-3)^2}+ \frac{m-3}{(2m-3)(2m+3)}= \\ \\ = \frac{2m(2m+3)+(m-3)(2m-3)}{(2m-3)^2(2m+3)}= \frac{4m^2+6m+2m^2-6m-3m+9}{(2m-3)^2(2m+3)}= \\ \\ = \frac{6m^2-3m+9}{(2m-3)^2(2m+3)}$

2)
$\frac{1-n}{25-9n^2}- \frac{3n}{(3n-5)^2}= \frac{n-1}{9n^2-25}- \frac{3n}{(3n-5)^2}= \\ \\ = \frac{n-1}{(3n-5)(3n+5)}- \frac{3n}{(3n-5)^2}= \frac{(n-1)(3n-5)-3n(3n+5)}{(3n-5)^2(3n+5)}= \\ \\ = \frac{3n^2-3n-5n+5-9n^2-15n}{(3n-5)^2(3n+5)}= \frac{-6n^2-23n+5}{(3n-5)^2(3n+5)}$