\(\eqalign{& {\left( {{{ - 3} \over 4}} \right)^2} = {{{{\left(- 3 \right)}^2}} \over {{4^2}}} = {9 \over {16}} \cr & {\left( {{{ - 2} \over 5}} \right)^3} = {{{{\left( { - 2} \right)}^3}} \over {{5^3}}} = {{ - 8} \over {125}} \cr& {\left( { - 0,5} \right)^2} = {\left( {{{ - 1} \over 2}} \right)^2} = {{{{\left( { - 1} \right)}^2}} \over {{2^2}}} = {1 \over 4} \cr & {\left( { - 0,5} \right)^3} = {\left( {{{ - 1} \over 2}} \right)^3} = {{{{\left( { - 1} \right)}^3}} \over {{2^3}}} = {{ - 1} \over 8} \cr & {\left( {9,7} \right)^0} = 1 \cr} \)
