# Express the following fraction in simplest form, only using positive exponents.

Express the following fraction in simplest form, only using positive exponents.
$$\displaystyle{\frac{{{20}{y}^{{2}}{h}^{{-{1}}}}}{{{\left({5}{y}^{{5}}{h}^{{2}}\right)}^{{2}}}}}$$

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