Упрощение алгебраических выражений. Задача 54

ОТВЕТ: \({\left( {\dfrac{m}{n}} \right)^{m + n}}\).

\(\dfrac{{{{\left( {{m^2}-{n^{-2}}} \right)}^m} \cdot {{\left( {n + {m^{-1}}} \right)}^{n-m}}}}{{{{\left( {{n^2}-{m^{-2}}} \right)}^n} \cdot {{\left( {m-{n^{-1}}} \right)}^{m-n}}}} = \dfrac{{{{\left( {{m^2}-\dfrac{1}{{{n^2}}}} \right)}^m} \cdot {{\left( {n + \dfrac{1}{m}} \right)}^{n-m}}}}{{{{\left( {{n^2}-\dfrac{1}{{{m^2}}}} \right)}^n} \cdot {{\left( {m-\dfrac{1}{n}} \right)}^{m-n}}}} = \)

\( = \dfrac{{{{\left( {\dfrac{{{m^2}{n^2}-1}}{{{n^2}}}} \right)}^m} \cdot {{\left( {\dfrac{{nm + 1}}{m}} \right)}^{n-m}}}}{{{{\left( {\dfrac{{{n^2}{m^2}-1}}{{{m^2}}}} \right)}^n} \cdot {{\left( {\dfrac{{mn-1}}{n}} \right)}^{m-n}}}} = \dfrac{{{{\left( {\dfrac{{{m^2}{n^2}-1}}{{{n^2}}}} \right)}^m} \cdot {{\left( {\dfrac{{nm + 1}}{m}} \right)}^{n-m}} \cdot {{\left( {\dfrac{{mn-1}}{n}} \right)}^{n-m}}}}{{{{\left( {\dfrac{{{n^2}{m^2}-1}}{{{m^2}}}} \right)}^n}}} = \)

\( = \dfrac{{{{\left( {\dfrac{{{m^2}{n^2}-1}}{{{n^2}}}} \right)}^m} \cdot {{\left( {\dfrac{{{m^2}{n^2}-1}}{{mn}}} \right)}^{n-m}}}}{{{{\left( {\dfrac{{{n^2}{m^2}-1}}{{{m^2}}}} \right)}^n}}} = \dfrac{{{{\left( {\dfrac{{{m^2}{n^2}-1}}{{{n^2}}}} \right)}^m} \cdot {{\left( {\dfrac{{{m^2}{n^2}-1}}{{mn}}} \right)}^n}}}{{{{\left( {\dfrac{{{m^2}{n^2}-1}}{{mn}}} \right)}^m} \cdot {{\left( {\dfrac{{{n^2}{m^2}-1}}{{{m^2}}}} \right)}^n}}} = {\left( {\dfrac{m}{n}} \right)^m} \cdot {\left( {\dfrac{m}{n}} \right)^n} = {\left( {\dfrac{m}{n}} \right)^{m + n}}.\)

Ответ:  \({\left( {\dfrac{m}{n}} \right)^{m + n}}.\)