\(x > 4\)
\(\dfrac{{\sqrt {x-4\sqrt {x-4} } + 2}}{{\sqrt {x + 4\sqrt {x-4} } -2}} = \dfrac{{\sqrt {x-4-4\sqrt {x-4} + 4} + 2}}{{\sqrt {x-4 + 4\sqrt {x-4} + 4} -2}} = \dfrac{{\sqrt {{{\left( {\sqrt {x-4} } \right)}^2}-4\sqrt {x-4} + {2^2}} + 2}}{{\sqrt {{{\left( {\sqrt {x-4} } \right)}^2} + 4\sqrt {x-4} + {2^2}} -2}} = \)
\( = \dfrac{{\sqrt {{{\left( {\sqrt {x-4} -2} \right)}^2}} + 2}}{{\sqrt {{{\left( {\sqrt {x-4} + 2} \right)}^2}} -2}} = \dfrac{{\left| {\sqrt {x-4} -2} \right| + 2}}{{\left| {\sqrt {x-4} + 2} \right|-2}} = \dfrac{{\left| {\sqrt {x-4} -2} \right| + 2}}{{\sqrt {x-4} + 2-2}} = \dfrac{{\left| {\sqrt {x-4} -2} \right| + 2}}{{\sqrt {x-4} }}.\)
\(\sqrt {x-4} -2 = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sqrt {x-4} = 2\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x = 8.\)
Если \(x\, \in \,\left( {4;8} \right)\), то
\(\dfrac{{\left| {\sqrt {x-4} -2} \right| + 2}}{{\sqrt {x-4} }} = \dfrac{{-\sqrt {x-4} + 2 + 2}}{{\sqrt {x-4} }} = \dfrac{4}{{\sqrt {x-4} }}-1.\)
Если \(x\, \in \,\left[ {8;\infty } \right)\), то
\(\dfrac{{\left| {\sqrt {x-4} -2} \right| + 2}}{{\sqrt {x-4} }} = \dfrac{{\sqrt {x-4} -2 + 2}}{{\sqrt {x-4} }} = \dfrac{{\sqrt {x-4} }}{{\sqrt {x-4} }} = 1.\)
Ответ: \(\dfrac{4}{{\sqrt {x-4} }}-1\), если \(x\, \in \,\left( {4;8} \right);\) 1, если \(x\, \in \,\left[ {8;\infty } \right)\).