Решение:

a)

$$\frac{5}{y-3} + \frac{1}{y+3} - \frac{4y-18}{y^2-9} =$$ $$\frac{5}{y-3} + \frac{1}{y+3} - \frac{4y-18}{(y-3)(y+3)} =$$ Приводим к общему знаменателю: (y-3)(y+3) $$\frac{5(y+3) + 1(y-3) - (4y-18)}{(y-3)(y+3)} =$$ $$\frac{5y + 15 + y - 3 - 4y + 18}{(y-3)(y+3)} =$$ $$\frac{2y + 30}{(y-3)(y+3)} =$$ $$\frac{2(y + 15)}{(y-3)(y+3)}$$ Ответ: $$\frac{2(y + 15)}{(y-3)(y+3)}$$

б)

$$\frac{2a}{2a+3} + \frac{5}{3-2a} - \frac{4a^2+9}{4a^2-9} =$$ $$\frac{2a}{2a+3} - \frac{5}{2a-3} - \frac{4a^2+9}{(2a+3)(2a-3)} =$$ Приводим к общему знаменателю: (2a+3)(2a-3) $$\frac{2a(2a-3) - 5(2a+3) - (4a^2+9)}{(2a+3)(2a-3)} =$$ $$\frac{4a^2 - 6a - 10a - 15 - 4a^2 - 9}{(2a+3)(2a-3)} =$$ $$\frac{-16a - 24}{(2a+3)(2a-3)} =$$ $$\frac{-8(2a + 3)}{(2a+3)(2a-3)} =$$ $$\frac{-8}{2a-3} =$$ $$\frac{8}{3-2a}$$ Ответ: $$\frac{8}{3-2a}$$

в)

$$\frac{4m}{4m^2-1} - \frac{2m+1}{6m-3} + \frac{2m-1}{4m+2} =$$ $$\frac{4m}{(2m-1)(2m+1)} - \frac{2m+1}{3(2m-1)} + \frac{2m-1}{2(2m+1)} =$$ Приводим к общему знаменателю: 6(2m-1)(2m+1) $$\frac{4m \cdot 6 - (2m+1) \cdot 2(2m+1) + (2m-1) \cdot 3(2m-1)}{6(2m-1)(2m+1)} =$$ $$\frac{24m - 2(4m^2 + 4m + 1) + 3(4m^2 - 4m + 1)}{6(2m-1)(2m+1)} =$$ $$\frac{24m - 8m^2 - 8m - 2 + 12m^2 - 12m + 3}{6(2m-1)(2m+1)} =$$ $$\frac{4m^2 + 4m + 1}{6(2m-1)(2m+1)} =$$ $$\frac{(2m+1)^2}{6(2m-1)(2m+1)} =$$ $$\frac{2m+1}{6(2m-1)}$$ Ответ: $$\frac{2m+1}{6(2m-1)}$$