\(\left( {x-1} \right)\sqrt {{x^2} + 2} > {x^2}-1\)
Запишем область определения: \({x^2} + 2 \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,R.\)
\(\left( {x-1} \right)\sqrt {{x^2} + 2} > {x^2}-1\,\,\,\,\, \Leftrightarrow \,\,\,\,\left( {x-1} \right)\sqrt {{x^2} + 2} -\left( {x-1} \right)\left( {x + 1} \right) > 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\left( {x-1} \right)\left( {\sqrt {{x^2} + 2} -\left( {x + 1} \right)} \right) > 0.\)
Решим данное неравенство методом интервалов:
\(\left( {x-1} \right)\left( {\sqrt {{x^2} + 2} -\left( {x + 1} \right)} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x-1 = 0,\\\sqrt {{x^2} + 2} -\left( {x + 1} \right) = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\\sqrt {{x^2} + 2} = x + 1\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\\left\{ \begin{array}{l}x + 1 \ge 0,\\{x^2} + 2 = {x^2} + 2x + 1\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\\left\{ \begin{array}{l}x \ge -1,\\x = 0,5\end{array} \right.\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 1,\\x = 0,5.\end{array} \right.\)
Ответ: \(\left( {\dfrac{1}{2};1} \right).\)