Решение:

a)

$$\frac{10p}{p-q} + \frac{3p}{q-p} = \frac{10p}{p-q} - \frac{3p}{p-q} = \frac{10p - 3p}{p-q} = \frac{7p}{p-q}$$

б)

$$\frac{5a}{a-b} + \frac{5b}{b-a} = \frac{5a}{a-b} - \frac{5b}{a-b} = \frac{5a - 5b}{a-b} = \frac{5(a - b)}{a-b} = 5$$

в)

$$\frac{x-3}{x-1} - \frac{2}{1-x} = \frac{x-3}{x-1} + \frac{2}{x-1} = \frac{x-3+2}{x-1} = \frac{x-1}{x-1} = 1$$

г)

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

д)

$$\frac{a}{a^2-9} + \frac{3}{9-a^2} = \frac{a}{a^2-9} - \frac{3}{a^2-9} = \frac{a - 3}{a^2-9} = \frac{a - 3}{(a - 3)(a + 3)} = \frac{1}{a + 3}$$

e)

$$\frac{y^2}{y-1} + \frac{1}{1-y} = \frac{y^2}{y-1} - \frac{1}{y-1} = \frac{y^2 - 1}{y-1} = \frac{(y - 1)(y + 1)}{y-1} = y + 1$$