\(\eqalign{ & a)x:{\left( { - {1 \over 2}} \right)^3} = - {1 \over 2} \cr & x = {{ - 1} \over 2}.{\left( {{{ - 1} \over 2}} \right)^3} \cr & x = {\left( { - {1 \over 2}} \right)^4} \cr & x = {1 \over {16}} \cr & b)x.{\left( {{3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^4} \cr & x = {\left( {{3 \over 5}} \right)^4}:{\left( {{3 \over 5}} \right)^3} \cr & x = {\left( {{3 \over 5}} \right)^{4 - 3}} \cr & x = {3 \over 5} \cr & c)x.{\left( { - {3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^6} \cr & x.{\left( { - {3 \over 5}} \right)^3} = {\left( {{{ - 3} \over 5}} \right)^6} \cr & x = {\left( { - {3 \over 5}} \right)^6}:{\left( {{{ - 3} \over 5}} \right)^3} \cr & x = {\left( {{{ - 3} \over 5}} \right)^3} \cr & x = {{ - 27} \over {125}} \cr} \) Đề bài Tìm x, biết a) \(x:{\left( { - {1 \over 2}} \right)^3} = - {1 \over 2}\) b) \(x.{\left( {{3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^4}\) c) \(x.{\left( { - {3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^6}\) Lời giải chi tiết \(\eqalign{ & a)x:{\left( { - {1 \over 2}} \right)^3} = - {1 \over 2} \cr & x = {{ - 1} \over 2}.{\left( {{{ - 1} \over 2}} \right)^3} \cr & x = {\left( { - {1 \over 2}} \right)^4} \cr & x = {1 \over {16}} \cr & b)x.{\left( {{3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^4} \cr & x = {\left( {{3 \over 5}} \right)^4}:{\left( {{3 \over 5}} \right)^3} \cr & x = {\left( {{3 \over 5}} \right)^{4 - 3}} \cr & x = {3 \over 5} \cr & c)x.{\left( { - {3 \over 5}} \right)^3} = {\left( {{3 \over 5}} \right)^6} \cr & x.{\left( { - {3 \over 5}} \right)^3} = {\left( {{{ - 3} \over 5}} \right)^6} \cr & x = {\left( { - {3 \over 5}} \right)^6}:{\left( {{{ - 3} \over 5}} \right)^3} \cr & x = {\left( {{{ - 3} \over 5}} \right)^3} \cr & x = {{ - 27} \over {125}} \cr} \)
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