\(2{\left( {xy} \right)^{\frac{1}{2}}}{\left( {x + y} \right)^{-1}}{\left( {1 + \dfrac{1}{4}{{\left( {\sqrt {\dfrac{x}{y}} -\sqrt {\dfrac{y}{x}} } \right)}^2}} \right)^{\frac{1}{2}}} = \)
\( = 2\sqrt {xy} \cdot \dfrac{1}{{x + y}} \cdot {\left( {1 + \dfrac{1}{4}\left( {\dfrac{x}{y}-2\sqrt {\dfrac{x}{y}} \cdot \sqrt {\dfrac{y}{x}} + \dfrac{y}{x}} \right)} \right)^{\frac{1}{2}}} = \)
\( = \dfrac{{2\sqrt {xy} }}{{x + y}}{\left( {1 + \dfrac{1}{4}\left( {\dfrac{x}{y}-2 + \dfrac{y}{x}} \right)} \right)^{\frac{1}{2}}} = \dfrac{{2\sqrt {xy} }}{{x + y}}{\left( {1 + \dfrac{1}{4} \cdot \dfrac{{{x^2}-2xy + {y^2}}}{{xy}}} \right)^{\frac{1}{2}}} = \)
\( = \dfrac{{2\sqrt {xy} }}{{x + y}} \cdot {\left( {\dfrac{{4xy + {x^2}-2xy + {y^2}}}{{4xy}}} \right)^{\frac{1}{2}}} = \dfrac{{2\sqrt {xy} }}{{x + y}} \cdot \sqrt {\dfrac{{{{\left( {x + y} \right)}^2}}}{{4xy}}} = \)
\( = \dfrac{{2\sqrt {xy} }}{{x + y}} \cdot \dfrac{{\left| {x + y} \right|}}{{2\sqrt {xy} }} = \dfrac{{\left| {x + y} \right|}}{{x + y}}.\)
Если \(x > 0,\,\,\,y > 0\), то \(\dfrac{{\left| {x + y} \right|}}{{x + y}} = \dfrac{{x + y}}{{x + y}} = 1.\)
Если \(x < 0,\,\,y < 0\), то \(\dfrac{{\left| {x + y} \right|}}{{x + y}} = \dfrac{{-\left( {x + y} \right)}}{{x + y}} = -1.\)
Ответ: 1, если \(x > 0,\,\,\,\,y > 0;\) \(-1\), если \(x < 0,\,\,\,\,y < 0.\)