\({1 \over a^5}=a^{-5}\)

\({a^2 \over a^{-9}}=a^{2--9}=a^{11}\)

\(x^2⋅x^{-5}=x^{2+-5}=x^{-3}\)

\((p^5)^{-2}=p^{5⋅-2}=p^{-10}\)

\(x^7⋅{1 \over x^8}=x^7⋅x^{-8}=x^{7+-8}=x^{-1}\)

\({({1 \over p^9}) \over p^2}={p^{-9} \over p^2}=p^{-9-2}=p^{-11}\)

\({x^0 \over x^9}=x^{0-9}=x^{-9}\)

\({6 \over a^9}\)

\({4a \over 7a^4}={4 \over 7}⋅{a \over a^4}={4 \over 7}⋅a^{1-4}={4 \over 7}a^{-3}\)

\({a^7 \over ({1 \over a^9})}={a^7 \over a^{-9}}=a^{7--9}=a^{16}\)

\({5x^3y^5 \over 3xy^8}={5 \over 3}⋅{x^3 \over x^1}⋅{y^5 \over y^8}={5 \over 3}⋅x^{3-1}⋅x^{5-8}=1\frac{2}{3}x^2y^{-3}\)

\(a^8⋅\sqrt[6]{a}=a^8⋅a^{\frac{1}{6}}=a^{8+\frac{1}{6}}=a^{8\frac{1}{6}}\)

\(a^2⋅\sqrt[9]{a^2}=a^2⋅a^{\frac{2}{9}}=a^{2+\frac{2}{9}}=a^{2\frac{2}{9}}\)

\({x^9 \over \sqrt[9]{x^5}}={x^9 \over x^{\frac{5}{9}}}=x^{9-\frac{5}{9}}=x^{8\frac{4}{9}}\)

\({1 \over p^5}⋅\sqrt[5]{p^2}=p^{-5}⋅p^{\frac{2}{5}}=p^{-5+\frac{2}{5}}=p^{-4\frac{3}{5}}\)

\({\sqrt[8]{a^7} \over \sqrt[4]{a^3}}={a^{\frac{7}{8}} \over a^{\frac{3}{4}}}=a^{\frac{7}{8}-\frac{3}{4}}=a^{\frac{1}{8}}\)

\(\sqrt[7]{{1 \over a^3}}=\sqrt[7]{a^{-3}}=a^{-\frac{3}{7}}\)

\(\sqrt[4]{x^{20}}=x^{\frac{20}{4}}=x^5\)

\({x^8 \over x^4⋅\sqrt[3]{x^2}}={x^8 \over x^4⋅x^{\frac{2}{3}}}={x^8 \over x^{4\frac{2}{3}}}=x^{8-4\frac{2}{3}}=x^{3\frac{1}{3}}\)

\({6y^9 \over 7x^3}\)

\((3p)^{-5}=3^{-5}⋅p^{-5}={1 \over 3^5}⋅{1 \over p^5}={1 \over 243p^5}\)

\(({1 \over 4}x)^{-3}=(4^{-1}⋅x)^{-3}=(4^{-1})^{-3}⋅x^{-3}=4^3⋅x^{-3}={64 \over x^3}\)

\(3p^{2\frac{7}{9}}=3⋅p^2⋅p^{\frac{7}{9}}=3p^2⋅\sqrt[9]{p^7}\)

\(\frac{3}{8}a^{-\frac{4}{9}}b^{\frac{3}{4}}=\frac{3}{8}⋅{1 \over a^{\frac{4}{9}}}⋅b^{\frac{3}{4}}={3⋅\sqrt[4]{b^3} \over 8⋅\sqrt[9]{a^4}}\)