Упр.6.12 ГДЗ Мордкович 8 класс (Алгебра)

а)

$$\left(\frac{10m^2}{3+2m}-5m\right):\frac{30m^2-15m}{8m^3+27}+\frac{8-2m}{2m-1}$$

$$\frac{10m^2-15m-10m^2}{3+2m}:\frac{15m(2m-1)}{(2m+3)(4m^2-6m+9)}+\frac{8-2m}{2m-1}$$

$$\frac{-15m}{3+2m}\cdot \frac{(2m+3)(4m^2-6m+9)}{15m(2m-1)}+\frac{8-2m}{2m-1}$$

$$-\frac{4m^2-6m+9}{2m-1}+\frac{8-2m}{2m-1} = \frac{-4m^2+6m-9+8-2m}{2m-1}$$

$$\frac{-4m^2+4m-1}{2m-1} = -\frac{(2m-1)^2}{2m-1} = -(2m-1)=1-2m$$

б)

$$\left(3n-\frac{9n^2}{3n+1}\right)\cdot \frac{27n^3+1}{6n-9n^2}+\frac{9n-3}{3n-2}$$

$$\frac{3n(3n+1)-9n^2}{3n+1}\cdot \frac{27n^3+1}{6n-9n^2}+\frac{9n-3}{3n-2} = \frac{9n^2+3n-9n^2}{3n+1}\cdot \frac{27n^3+1}{3n(2-3n)}+\frac{9n-3}{3n-2}$$

$$\frac{3n}{3n+1}\cdot \frac{(3n+1)(9n^2-3n+1)}{3n(2-3n)}+\frac{9n-3}{3n-2} = \frac{9n^2-3n+1}{2-3n}+\frac{9n-3}{3n-2}$$

$$\frac{9n^2-3n+1}{2-3n}-\frac{9n-3}{2-3n} = \frac{9n^2-12n+4}{2-3n} = \frac{(3n-2)^2}{2-3n} = 2-3n$$

Ответ

а) $$1-2m$$
б) $$2-3n$$