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