Упрощение алгебраических выражений. Задача 24

ОТВЕТ: \(2\left( {x + 1} \right)\).

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

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

\( = \dfrac{{{x^2} + 2xy + {y^2}}}{{y\left( {x-y} \right)}} \cdot \dfrac{{x\left( {x-y} \right) + \left( {x-y} \right)}}{{{x^2}}} \cdot \dfrac{{2{x^2}y}}{{{{\left( {x + y} \right)}^2}}} = \)

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

Ответ:  \(2\left( {x + 1} \right).\)