Упрощение алгебраических выражений. Задача 13

ОТВЕТ: \(\dfrac{{4 + {n^2}}}{{4n}}\).

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

\( = \dfrac{{2n-{n^2}-4 + 2n-{n^2}-4n-4}}{{\left( {n + 2} \right)\left( {n-2} \right)}} \cdot {\left( {\dfrac{{{n^2} + 4n + 4 + 2n-4-{n^2} + 2n}}{{\left( {2-n} \right)\left( {n + 2} \right)}}} \right)^{-1}} = \)

\( = \dfrac{{-2{n^2}-8}}{{\left( {n + 2} \right)\left( {n-2} \right)}} \cdot \dfrac{{-\left( {n-2} \right)\left( {n + 2} \right)}}{{8n}} = \dfrac{{2\left( {{n^2} + 4} \right)}}{{8n}} = \dfrac{{{n^2} + 4}}{{4n}}.\)

Ответ:  \(\dfrac{{4 + {n^2}}}{{4n}}.\)