\(a\, \in \,\left( {-1;1} \right).\)
\(\sqrt {{{\left( {1-a} \right)}^{-1}}} \,{\left( {\sqrt {1 + a} } \right)^{-1}}-\dfrac{{a\,\,\sqrt {{a^2}-{a^3}} \cdot \sqrt {1 + a} + {{\left( {\,1-{a^2}} \right)}^{\frac{1}{2}}}}}{{1-{a^4}}} = \)
\(\dfrac{1}{{\sqrt {1-a} \cdot \sqrt {1 + a} }}-\dfrac{{a\sqrt {{a^2}\left( {1-a} \right)} \cdot \sqrt {1 + a} + \sqrt {1-{a^2}} }}{{\left( {1-{a^2}} \right)\left( {1 + {a^2}} \right)}} = \)
\( = \dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{a \cdot \left| a \right| \cdot \sqrt {1-{a^2}} + \sqrt {1-{a^2}} }}{{\left( {1-{a^2}} \right)\left( {1 + {a^2}} \right)}} = \dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{\left( {a \cdot \left| a \right| + 1} \right)\sqrt {1-{a^2}} }}{{\left( {1-{a^2}} \right)\left( {1 + {a^2}} \right)}}.\)
Если \(a\, \in \,\left[ {0;1} \right)\), то \(\left| a \right| = a\):
\(\dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{\left( {{a^2} + 1} \right)\sqrt {1-{a^2}} }}{{\left( {1-{a^2}} \right)\left( {1 + {a^2}} \right)}} = \dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{\sqrt {1-{a^2}} }}{{{{\left( {\sqrt {1-{a^2}} } \right)}^2}}} = \dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{1}{{\sqrt {1-{a^2}} }} = 0.\)
Если \(a\, \in \,\left( {-1;0} \right)\), то \(\left| a \right| = -a\):
\(\dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{\left( {1-{a^2}} \right) \cdot \sqrt {1-{a^2}} }}{{\left( {1-{a^2}} \right) \cdot \left( {1 + {a^2}} \right)}} = \dfrac{1}{{\sqrt {1-{a^2}} }}-\dfrac{{\sqrt {1-{a^2}} }}{{1 + {a^2}}} = \)
\( = \dfrac{{1 + {a^2}-1 + {a^2}}}{{\sqrt {1-{a^2}} \cdot \left( {1 + {a^2}} \right)}} = \dfrac{{2{a^2}}}{{\sqrt {1-{a^2}} \cdot \left( {1 + {a^2}} \right)}}.\)
Ответ: 0, если \(a \in \left[ {0;1} \right);\) \(\dfrac{{2{a^2}}}{{\left( {1 + {a^2}} \right)\sqrt {1-{a^2}} }},\) если \(a \in \left( {-1;0} \right).\)