d+1d-4 r) 3d+d-2 =

Решение:

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

Ответ:

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