\(\dfrac{4}{{\sqrt {2-x} }}-\sqrt {2-x} < 2\)
Запишем область определения: \(2-x > 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x \in \left( {-\infty ;\,2} \right).\)
Пусть \(\sqrt {2-x} = t,\) где \(t > 0.\) Тогда:
\(\left\{ \begin{array}{l}\dfrac{4}{t}-t-2 < 0\\t > 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}{t^2} + 2t-4 > 0,\\t > 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ \begin{array}{l}\left[ \begin{array}{l}t < -1-\sqrt 5 ,\\t > -1 + \sqrt 5 \end{array} \right.\\t > 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,t > -1 + \sqrt 5 .\)
Возвращаясь к прежней переменной, получим:
\(\sqrt {2-x} > \sqrt 5 -1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,2-x > 5-2\sqrt 5 + 1\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x < 2\sqrt 5 -4.\)
Ответ: \(\left( {-\infty ;2\sqrt 5 -4} \right).\)