Negatieve en gebroken exponenten

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

\({2 a^{2} \over 3 a^{3}} = {2 \over 3} ⋅ {a^{2} \over a^{3}} = {2 \over 3} ⋅ a^{2 - 3} = {2 \over 3} a^{-1}\)

\({x^{2} \over x^{-4}} = x^{2 - -4} = x^{6}\)

\({({1 \over p^{5}}) \over p^{2}} = {p^{-5} \over p^{2}} = p^{-5 - 2} = p^{-7}\)

\({a^{7} \over ({1 \over a^{8}})} = {a^{7} \over a^{-8}} = a^{7 - -8} = a^{15}\)

\({5 a^{6} b^{5} \over 4 a b^{10}} = {5 \over 4} ⋅ {a^{6} \over a^{1}} ⋅ {b^{5} \over b^{10}} = {5 \over 4} ⋅ a^{6 - 1} ⋅ a^{5 - 10} = 1\frac{1}{4} a^{5} b^{-5}\)

\({x^{0} \over x^{6}} = x^{0 - 6} = x^{-6}\)

\((x^{5})^{-9} = x^{5 ⋅ -9} = x^{-45}\)

\(p^{3} ⋅ p^{-9} = p^{3 + -9} = p^{-6}\)

\(a^{5} ⋅ {1 \over a^{7}} = a^{5} ⋅ a^{-7} = a^{5 + -7} = a^{-2}\)

\(a^{5} ⋅ \sqrt[4]{a} = a^{5} ⋅ a^{\frac{1}{4}} = a^{5 + \frac{1}{4}} = a^{5\frac{1}{4}}\)

\(x^{8} ⋅ \sqrt[8]{x^{7}} = x^{8} ⋅ x^{\frac{7}{8}} = x^{8 + \frac{7}{8}} = x^{8\frac{7}{8}}\)

\({a^{2} \over \sqrt[5]{a^{3}}} = {a^{2} \over a^{\frac{3}{5}}} = a^{2 - \frac{3}{5}} = a^{1\frac{2}{5}}\)

\({1 \over x^{3}} ⋅ \sqrt[9]{x^{4}} = x^{-3} ⋅ x^{\frac{4}{9}} = x^{-3 + \frac{4}{9}} = x^{-2\frac{5}{9}}\)

\({\sqrt[9]{p^{4}} \over \sqrt[7]{p^{3}}} = {p^{\frac{4}{9}} \over p^{\frac{3}{7}}} = p^{\frac{4}{9} - \frac{3}{7}} = p^{\frac{1}{63}}\)

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

\(\sqrt[5]{p^{15}} = p^{\frac{15}{5}} = p^{3}\)

\({a^{2} \over a^{4} ⋅ \sqrt[9]{a^{8}}} = {a^{2} \over a^{4} ⋅ a^{\frac{8}{9}}} = {a^{2} \over a^{4\frac{8}{9}}} = a^{2 - 4\frac{8}{9}} = a^{-2\frac{8}{9}}\)

\((4 x)^{-5} = 4^{-5} ⋅ x^{-5} = {1 \over 4^{5}} ⋅ {1 \over x^{5}} = {1 \over 1\,024 x^{5}}\)

\(({1 \over 2} x)^{-3} = (2^{-1} ⋅ x)^{-3} = (2^{-1})^{-3} ⋅ x^{-3} = 2^{3} ⋅ x^{-3} = {8 \over x^{3}}\)

\({5 \over a^{4}}\)

\({5 q^{6} \over 8 p^{3}}\)

\(5 a^{4\frac{5}{7}} = 5 ⋅ a^{4} ⋅ a^{\frac{5}{7}} = 5 a^{4} ⋅ \sqrt[7]{a^{5}}\)

\(\frac{1}{8} x^{-\frac{2}{3}} y^{\frac{2}{7}} = \frac{1}{8} ⋅ {1 \over x^{\frac{2}{3}}} ⋅ y^{\frac{2}{7}} = {1 ⋅ \sqrt[7]{y^{2}} \over 8 ⋅ \sqrt[3]{x^{2}}}\)