Упр.6.4 ГДЗ Мордкович 8 класс (Алгебра)

а)
$$\left(2+\frac{t}{t+1}\right):\frac{3t^2+3t}{12t+8} = \frac{2t+2+t}{t+1}:\frac{3t(t+1)}{4(3t+2)}$$
$$= \frac{3t+2}{t+1}\cdot\frac{4(3t+2)}{3t(t+1)} = \frac{4(3t+2)^2}{3t(t+1)^2}.$$

б)
$$\left(p-\frac{5p}{p+2}\right):\frac{p-3}{p+2} = \frac{p(p+2)-5p}{p+2}:\frac{p-3}{p+2}$$
$$= \frac{p^2-3p}{p+2}\cdot\frac{p+2}{p-3} = \frac{p(p-3)(p+2)}{(p+2)(p-3)} = p.$$

в)
$$\frac{z-3}{z+3}\cdot\left(z+\frac{z^2}{3-z}\right) = \frac{z-3}{z+3}\cdot\frac{z(3-z)+z^2}{3-z}$$
$$= \frac{z-3}{z+3}\cdot\frac{3z-z^2+z^2}{3-z} = \frac{z-3}{z+3}\cdot\frac{3z}{3-z} = -\frac{3z}{z+3}.$$

г)
$$\left(\frac{q}{q-5}-2q\right):\frac{11-2q}{q-5} = \frac{q-2q(q-5)}{q-5}:\frac{11-2q}{q-5}$$
$$= \frac{q-2q^2+10q}{q-5}\cdot\frac{q-5}{11-2q} = \frac{11q-2q^2}{11-2q} = q.$$

Ответ

а) $$\frac{4(3t+2)^2}{3t(t+1)^2}$$; б) $$p$$; в) $$-\frac{3z}{z+3}$$; г) $$q$$.