Question

Запишите выражение в виде несократимой дроби без степеней с отрицательными показателями
$\frac{(5ab+a^{-3}) ^{2} }{5a^{4} b^{-2}+b^{-3} } -5a^{-2} b^{4}$

Answer 1

$\displaystyle \frac{(5ab+a^{-3})^2}{5a^4b^{-2}+b^{-3}}-5a^{-2}b^4=\frac{(a^{-3}(5a^4b+1))^2}{b^{-3}(5a^4b+1)}-5a^{-2}b^4=\\ \\ \\ =\frac{b^3(5a^4b+1)}{a^6}-\frac{5b^4}{a^2}=\frac{5a^4b^4+b^3-5a^4b^4}{a^6}=\dfrac{b^3}{a^6}$

Answer 2

$\frac{(5ab+a^{-3})^{2}}{5a^{4}b^{-2}+b^{-3}}-5a^{-2}b^{4}=\frac{(5ab+\frac{1}{a^{3}})^{2}}{5a^{4}*\frac{1}{b^{2} }+\frac{1}{b^{3}}}-5*\frac{1}{a^{2}}*b^{4}}=\frac{(\frac{5a^{4}b+1 }{a^{3}})^{2}}{\frac{5a^{4} }{b^{2}}+\frac{1}{b^{3}}}-\frac{5b^{4} }{a^{2} }=\frac{\frac{(5a^{4}b+1)^{2}}{a^{6} }}{\frac{5a^{4}b+1 }{b^{3} }}-\frac{5b^{4} }{a^{2} }=\frac{(5a^{4}b+1)^{2}*b^{3}}{(5a^{4}b+1)*a^{6}}-\frac{5b^{4} }{a^{2} }=\frac{(5a^{4}b+1)*b^{3}}{a^{6} }-\frac{5b^{4} }{a^{2} }=$$=\frac{5a^{4}b^{4}+b^{3}-5a^{4}b^{4}}{a^{6}}=\frac{b^{3} }{a^{6} }$