\(\sqrt {{x^2}-x + 1} < {\left( {x-1} \right)^2} + {x^2}\)
Запишем область определения: \({x^2}-x + 1 \ge 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,R.\)
\(\sqrt {{x^2}-x + 1} < {\left( {x-1} \right)^2} + {x^2}\,\,\,\, \Leftrightarrow \,\,\,\sqrt {{x^2}-x + 1} < 2{x^2}-2x + 1\,\,\,\, \Leftrightarrow \,\,\,\sqrt {{x^2}-x + 1} < 2\left( {{x^2}-x + 1} \right)-1.\)
Пусть \(\sqrt {{x^2}-x + 1} = t,\) где \(\,t \ge 0.\) Тогда: \({x^2}-x + 1 = {t^2}.\)
\(\left\{ \begin{array}{l}2{t^2}-t-1 > 0,\\t \ge 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}t < -\dfrac{1}{2},\\t > 1\end{array} \right.\\t \ge 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,t > 1.\)
Возвращаясь к прежней переменной, получим:
\(\sqrt {{x^2}-x + 1} > 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2}-x + 1 > 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,{x^2}-x > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;0} \right) \cup \left( {1;\infty } \right).\)
Ответ: \(\left( {-\infty ;0} \right) \cup \left( {1;\infty } \right).\)