Question

:3p^4/5q^8*15q^2 (p-5)^2/21p^2/3p^2 (5-p)/7q^6 должно получиться 5- p

Answer 1

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

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$\displaystyle (p-5)^2=p^2-10p+25=25-10p+p^2=(p-5)^5$