574. Сократите дробь: a) \(\frac{8a^3-27}{9-12a+4a^2}\); б) \(\frac{ax-2x-4a+8}{3a-6-ax+2x}\)

Ответ (RU):

1. a) $$\frac{8a^3-27}{9-12a+4a^2} = \frac{(2a-3)(4a^2+6a+9)}{(2a-3)^2} = \frac{4a^2+6a+9}{2a-3}$$
2. б) $$\frac{ax-2x-4a+8}{3a-6-ax+2x} = \frac{x(a-2)-4(a-2)}{3(a-2)-x(a-2)} = \frac{(a-2)(x-4)}{(a-2)(3-x)} = \frac{x-4}{3-x}$$

Перевод (EN):

1. a) $$\frac{8a^3-27}{9-12a+4a^2} = \frac{(2a-3)(4a^2+6a+9)}{(2a-3)^2} = \frac{4a^2+6a+9}{2a-3}$$
2. b) $$\frac{ax-2x-4a+8}{3a-6-ax+2x} = \frac{x(a-2)-4(a-2)}{3(a-2)-x(a-2)} = \frac{(a-2)(x-4)}{(a-2)(3-x)} = \frac{x-4}{3-x}$$

Ответ: a) \(\frac{4a^2+6a+9}{2a-3}\); б) \(\frac{x-4}{3-x}\)