\(\,x \ne 0,\,\,\,\,x \ne -1.\)
\(\sqrt {\dfrac{x}{{2 + x + {x^{-1}}}}} + \left| {x-1} \right| = \sqrt {\dfrac{x}{{2 + x + \dfrac{1}{x}}}} + \left| {x-1} \right| = \sqrt {\dfrac{{{x^2}}}{{{x^2} + 2x + 1}}} + \left| {x-1} \right| = \)
\( = \sqrt {{{\left( {\dfrac{x}{{x + 1}}} \right)}^2}} + \left| {x-1} \right| = \dfrac{{\left| x \right|}}{{\left| {x + 1} \right|}} + \left| {x-1} \right|.\)
Если \(x\, \in \,\left( {-\infty ;-1} \right)\), то
\(\dfrac{{\left| x \right|}}{{\left| {x + 1} \right|}} + \left| {x-1} \right| = \dfrac{{-x}}{{-\left( {x + 1} \right)}}-\left( {x-1} \right) = \dfrac{{x-{x^2} + 1}}{{x + 1}}.\)
Если \(x\, \in \,\left( {-1;0} \right)\), то
\(\dfrac{{\left| x \right|}}{{\left| {x + 1} \right|}} + \left| {x-1} \right| = \dfrac{{-x}}{{x + 1}}-\left( {x-1} \right) = \dfrac{{-x-{x^2} + 1}}{{x + 1}}.\)
Если \(x\, \in \,\left( {0;1} \right)\), то
\(\dfrac{{\left| x \right|}}{{\left| {x + 1} \right|}} + \left| {x-1} \right| = \dfrac{x}{{x + 1}}-\left( {x-1} \right) = \dfrac{{x-{x^2} + 1}}{{x + 1}}.\)
Если \(x\, \in \,\left[ {1;\infty } \right)\), то
\(\dfrac{{\left| x \right|}}{{\left| {x + 1} \right|}} + \left| {x-1} \right| = \dfrac{x}{{x + 1}} + \left( {x-1} \right) = \dfrac{{{x^2} + x-1}}{{x + 1}}.\)
Ответ: \(\dfrac{{1 + x-{x^2}}}{{x + 1}}\), если \(x\, \in \left( {-\infty ;-1} \right) \cup \left( {0;1} \right);\) \(\dfrac{{1-x-{x^2}}}{{x + 1}}\), если \(x\, \in \left( {-1;0} \right);\) \(\dfrac{{{x^2} + x-1}}{{x + 1}}\), если \(x\, \in \left[ {1;\infty } \right)\).