We have to simplify the following expression into simplest form using only positive exponents.

\(\displaystyle{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}\)

we have

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}\right)}^{{2}}{\left({y}^{{5}}\right)}^{{2}}{\left({h}^{{2}}\right)}^{{2}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{25}{y}^{{{10}}}{h}^{{4}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}{y}^{{2}}}}{{{5}{y}^{{{10}}}{h}^{{4}}{h}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}}}{{{5}{y}^{{8}}{h}^{{5}}}}}\)

Which is the required simplest form.

Hence we get,

\(\displaystyle{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}={\frac{{{4}}}{{{5}{y}^{{8}}{h}^{{5}}}}}\)

\(\displaystyle{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}\)

we have

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}\right)}^{{2}}{\left({y}^{{5}}\right)}^{{2}}{\left({h}^{{2}}\right)}^{{2}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{25}{y}^{{{10}}}{h}^{{4}}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}{y}^{{2}}}}{{{5}{y}^{{{10}}}{h}^{{4}}{h}}}}\)

\(\displaystyle\Rightarrow{\frac{{{4}}}{{{5}{y}^{{8}}{h}^{{5}}}}}\)

Which is the required simplest form.

Hence we get,

\(\displaystyle{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}={\frac{{{4}}}{{{5}{y}^{{8}}{h}^{{5}}}}}\)