Phương pháp giải:

Chuyển vế rồi quy đồng.

\(\dfrac{{f\left( x \right)}}{{\sqrt {g\left( x \right)} }} \le 0 \Leftrightarrow \left\{ \begin{array}{l}f\left( x \right) \le 0\\g\left( x \right) > 0\end{array} \right.\)

Lời giải chi tiết:

\(\begin{array}{l}f'\left( x \right) = \sqrt {{x^2} - 1}  + \left( {x - 2} \right).\dfrac{x}{{\sqrt {{x^2} - 1} }}\\ = \dfrac{{\left( {{x^2} - 1} \right) + \left( {x - 2} \right).x}}{{\sqrt {{x^2} - 1} }}\\ = \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }}\\f'\left( x \right) \le \sqrt {{x^2} - 1} \\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }} \le \sqrt {{x^2} - 1} \\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }} - \sqrt {{x^2} - 1}  \le 0\\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1 - \left( {{x^2} - 1} \right)}}{{\sqrt {{x^2} - 1} }} \le 0\\ \Leftrightarrow \dfrac{{{x^2} - 2x}}{{\sqrt {{x^2} - 1} }} \le 0 \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 2x \le 0\\{x^2} - 1 > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}0 \le x \le 2\\\left[ \begin{array}{l}x > 1\\x <  - 1\end{array} \right.\end{array} \right. \Leftrightarrow 1 < x \le 2\\ =  > S = \left( {1;2} \right]\end{array}\)