Решение:

a) $$rac{a+3}{a^2-3a} + \frac{a-9}{3a-9} = \frac{a+3}{a(a-3)} + \frac{a-9}{3(a-3)} = \frac{3(a+3) + a(a-9)}{3a(a-3)} = \frac{3a+9 + a^2-9a}{3a(a-3)} = \frac{a^2 - 6a + 9}{3a(a-3)} = \frac{(a-3)^2}{3a(a-3)} = \frac{a-3}{3a}$$

б) $$rac{b-4}{2b-4} - \frac{2}{2b-b^2} = \frac{b-4}{2(b-2)} - \frac{2}{b(2-b)} = \frac{b-4}{2(b-2)} + \frac{2}{b(b-2)} = \frac{b(b-4) + 2*2}{2b(b-2)} = \frac{b^2 - 4b + 4}{2b(b-2)} = \frac{(b-2)^2}{2b(b-2)} = \frac{b-2}{2b}$$

в) $$rac{2}{x^2+2xy} + \frac{1}{xy+2y^2} = \frac{2}{x(x+2y)} + \frac{1}{y(x+2y)} = \frac{2y + x}{xy(x+2y)} = \frac{x+2y}{xy(x+2y)} = \frac{1}{xy}$$

г) $$rac{1}{xy-x^2} - \frac{1}{y^2-xy} = \frac{1}{x(y-x)} - \frac{1}{y(y-x)} = \frac{y - x}{xy(y-x)} = \frac{-(x-y)}{xy(y-x)} = -\frac{1}{xy}$$

д) $$rac{c}{c^2-9} + \frac{c+2}{c^2-3c} = \frac{c}{(c-3)(c+3)} + \frac{c+2}{c(c-3)} = \frac{c^2 + (c+2)(c+3)}{c(c-3)(c+3)} = \frac{c^2 + c^2 + 3c + 2c + 6}{c(c-3)(c+3)} = \frac{2c^2 + 5c + 6}{c(c-3)(c+3)}$$

е) $$rac{d+1}{d+4} - \frac{d^2-4}{d^2-16} = \frac{d+1}{d+4} - \frac{(d-2)(d+2)}{(d-4)(d+4)} = \frac{(d+1)(d-4) - (d-2)(d+2)}{(d+4)(d-4)} = \frac{d^2 - 4d + d - 4 - (d^2 - 4)}{(d+4)(d-4)} = \frac{d^2 - 3d - 4 - d^2 + 4}{(d+4)(d-4)} = \frac{-3d}{(d+4)(d-4)}$$