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