Решение заданий:

1. $$\frac{1}{(a+b)^2} + \frac{1}{a+b} = \frac{1}{(a+b)^2} + \frac{a+b}{(a+b)^2} = \frac{1+a+b}{(a+b)^2}$$
2. $$\frac{1}{a-b} + \frac{1}{(a-b)^2} = \frac{a-b}{(a-b)^2} + \frac{1}{(a-b)^2} = \frac{a-b+1}{(a-b)^2}$$
3. $$\frac{1}{x-y} - \frac{1}{(x-y)^2} = \frac{x-y}{(x-y)^2} - \frac{1}{(x-y)^2} = \frac{x-y-1}{(x-y)^2}$$
4. $$\frac{3}{(n-m)^2} + \frac{1}{n-m} = \frac{3}{(n-m)^2} + \frac{n-m}{(n-m)^2} = \frac{3+n-m}{(n-m)^2}$$
5. $$\frac{2}{x+y} - \frac{a}{(x+y)^2} = \frac{2(x+y)}{(x+y)^2} - \frac{a}{(x+y)^2} = \frac{2x+2y-a}{(x+y)^2}$$
6. $$\frac{5}{2a+b} - \frac{x}{(2a+b)^2} = \frac{5(2a+b)}{(2a+b)^2} - \frac{x}{(2a+b)^2} = \frac{10a+5b-x}{(2a+b)^2}$$
7. $$\frac{1}{(m-2n)^2} - \frac{1}{(m-2n)^3} = \frac{m-2n}{(m-2n)^3} - \frac{1}{(m-2n)^3} = \frac{m-2n-1}{(m-2n)^3}$$
8. $$\frac{3}{2(x+y)} + \frac{b}{(x+y)^2} = \frac{3(x+y)}{2(x+y)^2} + \frac{2b}{2(x+y)^2} = \frac{3x+3y+2b}{2(x+y)^2}$$
9. $$\frac{x}{(a-b)^2} + \frac{2}{5(a-b)} = \frac{5x}{5(a-b)^2} + \frac{2(a-b)}{5(a-b)^2} = \frac{5x+2a-2b}{5(a-b)^2}$$
10. $$\frac{b}{(x-y)^3} + \frac{10}{3(x-y)^2} = \frac{3b}{3(x-y)^3} + \frac{10(x-y)}{3(x-y)^3} = \frac{3b+10x-10y}{3(x-y)^3}$$