\(\dfrac{{1-\sqrt {1-8{x^2}} }}{{2x}} < 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\dfrac{{1-\sqrt {1-8{x^2}} -2x}}{{2x}} < 0.\)
Решим данное неравенство методом интервалов при условии, что:
\(1-8{x^2} \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left[ {-\dfrac{1}{{2\sqrt 2 }};\,\dfrac{1}{{2\sqrt 2 }}} \right].\)
Найдём нули числителя:
\(1-\sqrt {1-8{x^2}} -2x = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\sqrt {1-8{x^2}} = 1-2x\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{1-8{x^2} = {{\left( {1-2x} \right)}^2},}\\{1-2x \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}12{x^2}-4x = 0,\\x \le \dfrac{1}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}x = 0,\\x = \dfrac{1}{3},\end{array} \right.\\x \le \dfrac{1}{2}\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = 0,\\x = \dfrac{1}{3}.\end{array} \right.\)
Найдём нули знаменателя: \(x = 0.\)
Ответ: \(\,\left[ {-\dfrac{1}{{2\sqrt 2 }};0} \right) \cup \left( {0;\dfrac{1}{3}} \right).\)