\(\dfrac{{\sqrt {1-{x^2}} -1}}{x}\left( {\dfrac{{1-x}}{{\sqrt {1-{x^2}} + x-1}} + \dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} -\sqrt {1-x} }}} \right) = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \left( {\dfrac{{{{\left( {\sqrt {1-x} } \right)}^2}}}{{\sqrt {1-x} \sqrt {1 + x} -{{\left( {\sqrt {1-x} } \right)}^2}}} + \dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} -\sqrt {1-x} }}} \right) = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \left( {\dfrac{{{{\left( {\sqrt {1-x} } \right)}^2}}}{{\sqrt {1-x} \left( {\sqrt {1 + x} -\sqrt {1-x} } \right)}} + \dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} -\sqrt {1-x} }}} \right) = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \left( {\dfrac{{\sqrt {1-x} }}{{\sqrt {1 + x} -\sqrt {1-x} }} + \dfrac{{\sqrt {1 + x} }}{{\sqrt {1 + x} -\sqrt {1-x} }}} \right) = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \dfrac{{\left( {\sqrt {1-x} + \sqrt {1 + x} } \right)\left( {\sqrt {1-x} + \sqrt {1-x} } \right)}}{{\left( {\sqrt {1 + x} -\sqrt {1-x} } \right)\left( {\sqrt {1 + x} + \sqrt {1-x} } \right)}} = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \dfrac{{1-x + 2\sqrt {1-x} \sqrt {1 + x} + 1 + x}}{{1 + x-1 + x}} = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \dfrac{{2 + 2\sqrt {1-{x^2}} }}{{2x}} = \)
\( = \dfrac{{\sqrt {1-{x^2}} -1}}{x} \cdot \dfrac{{1 + \sqrt {1-{x^2}} }}{x} = \dfrac{{{{\left( {\sqrt {1-{x^2}} } \right)}^2}-{1^2}}}{{{x^2}}} = \dfrac{{1-{x^2}-1}}{{{x^2}}} = -\dfrac{{{x^2}}}{{{x^2}}} = -1.\)
Ответ: \(-1\), если \(x\, \in \,\left[ {-1;0} \right) \cup \left( {0;1} \right].\)