Вопрос в картинках...

$\frac{3p^4}{5q^8}* \frac{15q^2*(p-5)^2}{21p^2} : \frac{2p^2(5-p)}{7q^6}= \frac{3p^4}{5q^8}* \frac{15q^2*(p-5)^2}{21p^2} * \frac{7q^6}{2p^2(5-p)}= \\ \frac{p^4}{q^8}* \frac{3q^2*(p-5)^2}{7p^2} * \frac{7q^6}{2p^2(5-p)}=\frac{p^4}{q^8}* \frac{3q^2*(5-p)^2}{p^2} * \frac{q^6}{2p^2(5-p)}= \\ \frac{p^2}{q^8}* \frac{3q^2*(5-p)}{1} * \frac{q^6}{2p^2}=\frac{p^2}{q^6}* \frac{3*1*(5-p)}{1} * \frac{q^6}{2p^2}= \\ =\frac{p^2}{1}* \frac{3*(5-p)}{1} * \frac{1}{2p^2}=\frac{1}{1}* \frac{3*(5-p)}{1} * \frac{1}{2*1}=$
$= \frac{3(5-p)}{2} =\frac{15-3p}{2} =7.5-1.5p$

примечание: (p-5)²=(5-p)²