Упрощение алгебраических выражений. Задача 7

ОТВЕТ: \(\dfrac{1}{{4x}}\).

\(\dfrac{2}{{4x-{x^2}}} + \left( {\dfrac{1}{{{x^2}-4x}} + \dfrac{2}{{16-{x^2}}} + \dfrac{1}{{16 + 4x}}} \right)\,{\left( {\dfrac{{x + 4}}{{x-4}}} \right)^2} = \)

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

\( = \dfrac{2}{{4x-{x^2}}} + \dfrac{{4x + 16-8x + {x^2}-4x}}{{4x\left( {x-4} \right)\left( {x + 4} \right)}} \cdot \dfrac{{{{\left( {x + 4} \right)}^2}}}{{{{\left( {x-4} \right)}^2}}} = \dfrac{2}{{4x-{x^2}}} + \dfrac{{\left( {{x^2}-8x + 16} \right)\left( {x + 4} \right)}}{{4x\left( {x-4} \right){{\left( {x-4} \right)}^2}}} = \)

\( = \dfrac{2}{{4x-{x^2}}} + \dfrac{{{{\left( {x-4} \right)}^2}\left( {x + 4} \right)}}{{4x\left( {x-4} \right){{\left( {x-4} \right)}^2}}} = \dfrac{{-2}}{{x\left( {x-4} \right)}} + \dfrac{{x + 4}}{{4x\left( {x-4} \right)}} = \dfrac{{-8 + x + 4}}{{4x\left( {x-4} \right)}} = \dfrac{{x-4}}{{4x\left( {x-4} \right)}} = \dfrac{1}{{4x}}.\)

Ответ:  \(\dfrac{1}{{4x}}.\)