\(\dfrac{{\left( {x-1} \right)\sqrt {{{\left( {x-1} \right)}^2} + 4x} }}{{{x^2} + 1 + 2\left| x \right|}} = \dfrac{{\left( {x-1} \right)\sqrt {{x^2}-2x + 1 + 4x} }}{{{{\left| x \right|}^2} + 2\left| x \right| + 1}} = \)
\( = \dfrac{{\left( {x-1} \right)\sqrt {{x^2} + 2x + 1} }}{{{{\left( {\left| x \right| + 1} \right)}^2}}} = \dfrac{{\left( {x-1} \right)\sqrt {{{\left( {x + 1} \right)}^2}} }}{{{{\left( {\left| x \right| + 1} \right)}^2}}} = \dfrac{{\left( {x-1} \right)\left| {x + 1} \right|}}{{{{\left( {\left| x \right| + 1} \right)}^2}}}.\)
Если \(x\, \in \,\left( {-\infty ;-1} \right)\), то
\(\dfrac{{\left( {x-1} \right)\left| {x + 1} \right|}}{{{{\left( {\left| x \right| + 1} \right)}^2}}} = \dfrac{{-\left( {x-1} \right)\left( {x + 1} \right)}}{{{{\left( {1-x} \right)}^2}}} = \dfrac{{\left( {1-x} \right)\left( {x + 1} \right)}}{{{{\left( {1-x} \right)}^2}}} = \dfrac{{x + 1}}{{1-x}}.\)
Если \(x\, \in \,\left[ {-1;0} \right)\), то
\(\dfrac{{\left( {x-1} \right)\left| {x + 1} \right|}}{{{{\left( {\left| x \right| + 1} \right)}^2}}} = \dfrac{{\left( {x-1} \right)\left( {x + 1} \right)}}{{{{\left( {1-x} \right)}^2}}} = \dfrac{{x + 1}}{{x-1}}.\)
Если \(x\, \in \,\left[ {0;\infty } \right)\), то
\(\dfrac{{\left( {x-1} \right)\left| {x + 1} \right|}}{{{{\left( {\left| x \right| + 1} \right)}^2}}} = \dfrac{{\left( {x-1} \right)\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}} = \dfrac{{x-1}}{{x + 1}}.\)
Ответ: \(\dfrac{{x + 1}}{{1-x}}\), если \(x\, \in \left( {-\infty ;-1} \right);\) \(\dfrac{{x + 1}}{{x-1}}\), если \(x\, \in \left[ {-1;0} \right);\) \(\dfrac{{x-1}}{{x + 1}}\), если \(x\, \in \left[ {0;\infty } \right)\).