Решение:

a)

$$\frac{c+1}{c-3} - \frac{2c}{c+3} = \frac{(c+1)(c+3) - 2c(c-3)}{(c-3)(c+3)} = \frac{c^2 + 3c + c + 3 - 2c^2 + 6c}{c^2 - 9} = \frac{-c^2 + 10c + 3}{c^2 - 9}$$

б)

$$\frac{y+1}{2y} - \frac{y-2}{y-3} = \frac{(y+1)(y-3) - (y-2)(2y)}{2y(y-3)} = \frac{y^2 - 3y + y - 3 - 2y^2 + 4y}{2y^2 - 6y} = \frac{-y^2 + 2y - 3}{2y^2 - 6y}$$

в)

$$\frac{b}{3b-2} - \frac{b+1}{3b+2} = \frac{b(3b+2) - (b+1)(3b-2)}{(3b-2)(3b+2)} = \frac{3b^2 + 2b - (3b^2 - 2b + 3b - 2)}{9b^2 - 4} = \frac{3b^2 + 2b - 3b^2 - b + 2}{9b^2 - 4} = \frac{b + 2}{9b^2 - 4}$$

г)

$$\frac{d+1}{3d} + \frac{d-4}{d-2} = \frac{(d+1)(d-2) + (d-4)(3d)}{3d(d-2)} = \frac{d^2 - 2d + d - 2 + 3d^2 - 12d}{3d^2 - 6d} = \frac{4d^2 - 13d - 2}{3d^2 - 6d}$$

Ответ:

a) $$\frac{-c^2 + 10c + 3}{c^2 - 9}$$

б) $$\frac{-y^2 + 2y - 3}{2y^2 - 6y}$$

в) $$\frac{b + 2}{9b^2 - 4}$$

г) $$\frac{4d^2 - 13d - 2}{3d^2 - 6d}$$