Помогите решить! 1)(z+6/ 3z+9 - 1/ z+3) : z+2/ 27z 2)m-2/ m-5 : (m²+24/ m²-25 - 4/ m-5)

Решите задачу:

$(\frac{z+6}{3z+9}- \frac{1}{z+3}): \frac{z+2}{27z}=( \frac{z+6}{3(z+3)}-\frac{1}{z+3}})* \frac{27z}{z+2}= \\ \\ \frac{z+6-3}{3(z+3)}* \frac{27z}{z+2}= \frac{z+3}{3(z+3)}* \frac{27z}{z+2}= \frac{9z}{z+2}$

$\frac{m-2}{m-5}:( \frac{m^2+24}{m^2-25y}- \frac{4}{m-5})= \frac{m-5}{m-2}*( \frac{m^2+24}{(m-5)(m+5)}- \frac{4}{m-5})= \\ \\ \frac{m-5}{m-2}* \frac{m^2+24-4(m+5)}{(m-5)(m+5)}= \frac{m^2+24-4m-20}{(m-2)(m+5)}= \frac{m^2-4m+4}{(m-2)(m+5)}= \\ \\ \frac{(m-2)^2}{(m-2)(m+5)}= \frac{m-2}{m+5}$