Phương pháp giải:

+) Tính \(f'\left( x \right)\).

+) Giải BPT dạng \(\sqrt {f\left( x \right)}  > g\left( x \right) \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) \ge 0\\g\left( x \right) < 0\end{array} \right.\\\left\{ \begin{array}{l}g\left( x \right) \ge 0\\f\left( x \right) > {g^2}\left( x \right)\end{array} \right.\end{array} \right.\) .

Lời giải chi tiết:

\(\begin{array}{l}DKXD:\,\, - 1 \le x \le 1\\f'\left( x \right) = 2 + \dfrac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} = 2 - \dfrac{x}{{\sqrt {1 - {x^2}} }}\\f'\left( x \right) > 0 \Leftrightarrow 2 - \dfrac{x}{{\sqrt {1 - {x^2}} }} > 0 \Leftrightarrow \dfrac{{2\sqrt {1 - {x^2}}  - x}}{{\sqrt {1 - {x^2}} }} > 0\,\,\left( {x \in \left( { - 1;1} \right)} \right)\\ \Leftrightarrow 2\sqrt {1 - {x^2}}  - x > 0 \Leftrightarrow 2\sqrt {1 - {x^2}}  > x\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}1 - {x^2} > 0\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\4\left( {1 - {x^2}} \right) > {x^2}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} - 1 < x < 1\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\5{x^2} < 4\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\\left\{ \begin{array}{l}x \ge 0\\\dfrac{{ - 2}}{{\sqrt 5 }} < x < \dfrac{2}{{\sqrt 5 }}\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 1 < x < 0\\0 \le x < \dfrac{2}{{\sqrt 5 }}\end{array} \right. \Leftrightarrow  - 1 < x < \dfrac{2}{{\sqrt 5 }}\,\,\left( {tm\,\,DKXD} \right)\end{array}\)

Vậy nghiệm của BPT là:  \(x \in \left( { - 1;\dfrac{2}{{\sqrt 5 }}} \right).\)

Chọn C.