\(x \ne 2\,,\,\,\,\,\,x \ne -2 \pm 2\sqrt 2 .\)
\(\dfrac{{\left( {x + 2} \right)\sqrt {{{\left( {x + 2} \right)}^2}-8x} }}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{\left( {x + 2} \right)\sqrt {{x^2} + 4x + 4-8x} }}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{\left( {x + 2} \right)\sqrt {{{\left( {x-2} \right)}^2}} }}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{\left( {x + 2} \right)\left| {x-2} \right|}}{{{x^2}-4\left| {x-1} \right|}}.\)
Если \(x\, \in \,\left( {-\infty ;-2-2\sqrt 2 } \right) \cup \left( {-2-2\sqrt 2 ;-2 + 2\sqrt 2 } \right) \cup \left( {-2 + 2\sqrt 2 ;1} \right)\), то
\(\dfrac{{\left( {x + 2} \right)\left| {x-2} \right|}}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{-\left( {x + 2} \right)\left( {x-2} \right)}}{{{x^2} + 4x-4}} = \dfrac{{4-{x^2}}}{{{x^2} + 4x-4}}.\)
Если \(x\, \in \,\left[ {1;2} \right)\), то
\(\dfrac{{\left( {x + 2} \right)\left| {x-2} \right|}}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{-\left( {x + 2} \right)\left( {x-2} \right)}}{{{x^2}-4x + 4}} = \dfrac{{-\left( {x + 2} \right)\left( {x-2} \right)}}{{{{\left( {x-2} \right)}^2}}} = \dfrac{{x + 2}}{{2-x}}.\)
Если \(x\, \in \,\left( {2;\infty } \right)\), то
\(\dfrac{{\left( {x + 2} \right)\left| {x-2} \right|}}{{{x^2}-4\left| {x-1} \right|}} = \dfrac{{\left( {x + 2} \right)\left( {x-2} \right)}}{{{x^2}-4x + 4}} = \dfrac{{\left( {x + 2} \right)\left( {x-2} \right)}}{{{{\left( {x-2} \right)}^2}}} = \dfrac{{x + 2}}{{x-2}}.\)
Ответ: \(\dfrac{{4-{x^2}}}{{{x^2} + 4x-4}}\), если \(x\, \in \,\left( {-\infty ;-2-2\sqrt 2 } \right) \cup \left( {-2-2\sqrt 2 ;-2 + 2\sqrt 2 } \right) \cup \left( {-2 + 2\sqrt 2 ;1} \right);\) \(\dfrac{{x + 2}}{{2-x}}\), если \(x\, \in \left[ {1;2} \right);\) \(\dfrac{{x + 2}}{{x-2}}\), если \(x\, \in \left( {2;\infty } \right)\).