Đáp án đúng: C

Phương pháp giải:

Thiện hiện quy đồng mẫu số sau đó áp dụng:\(\left| A \right| \ge B \Leftrightarrow \left[ \begin{array}{l}A \ge B\\A \le  - B\end{array} \right.\); \(\left| A \right| \le B \Leftrightarrow  - B \le A \le B\)Giải chi tiết:\(\begin{array}{l}\dfrac{{\left| {{x^2} - 4x} \right| + 3}}{{{x^2} + \left| {x - 5} \right|}} \ge 1\\ \Leftrightarrow \left| {{x^2} - 4x} \right| + 3 \ge {x^2} + \left| {x - 5} \right|\left( {{\mathop{\rm vì}\nolimits} \,\,{x^2} + \left| {x - 5} \right| \ge 0,\forall x \in \mathbb{R}} \right)\\ \Leftrightarrow \left| {{x^2} - 4x} \right| \ge {x^2} - 3 + \left| {x - 5} \right|\\ \Leftrightarrow \left[ \begin{array}{l}{x^2} - 4x \ge {x^2} - 3 + \left| {x - 5} \right|\\{x^2} - 4x \le  - {x^2} + 3 - \left| {x - 5} \right|\end{array} \right.\end{array}\)\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3 - 4x \ge \left| {x - 5} \right|\\2{x^2} - 4x - 3 \le  - \left| {x - 5} \right|\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left| {x - 5} \right| \le 3 - 4x\\\left| {x - 5} \right| \le  - 2{x^2} + 4x + 3\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 3 + 4x \le x - 5 \le 3 - 4x\\2{x^2} - 4x - 3 \le x - 5 \le  - 2{x^2} + 4x + 3\end{array} \right.\\\end{array}\)\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l} - 3 + 4x \le x - 5\\x - 5 \le 3 - 4x\end{array} \right.\\\left\{ \begin{array}{l}2{x^2} - 4x - 3 \le x - 5\\x - 5 \le  - 2{x^2} + 4x + 3\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x \le  - \dfrac{2}{3}\\\dfrac{1}{2} \le x \le 2\end{array} \right.\end{array}\)Vậy \(S = \left( { - \infty ;\,\, - \dfrac{2}{3}} \right] \cup \left[ {\dfrac{1}{2};\,\,2} \right]\).Đáp án C.