\(f\left( x \right) = \left| {\frac{{{x^2} + \left( {m - 2} \right)x + 2 - m}}{{x - 1}}} \right| = \left| {\frac{{{x^2} - 2x + 2}}{{x - 1}} + m} \right|\).

Xét hàm số \(g\left( x \right) = \frac{{{x^2} - 2x + 2}}{{x - 1}}\) trên đoạn [2;3], ta có

\(g'\left( x \right) = \frac{{{x^2} - 2x}}{{{{\left( {x - 1} \right)}^2}}} \ge 0,\,\forall x \in \left[ {2\,;\,3} \right]\)(g'(x) = 0 tại x = 2). Suy ra, tập giá trị của g(x) trên [2;3] là đoạn \(\left[ {g\left( 2 \right)\,;\,g\left( 3 \right)} \right] = \left[ {2\,;\,\frac{5}{2}} \right]\).

Đặt \(t = \frac{{{x^2} - 2x + 2}}{{x - 1}}\), hàm số f(x) trên [2;3] trở thành hàm số \(h\left( t \right) = \left| {t + m} \right|\) xét trên \(\left[ {2\,;\,\frac{5}{2}} \right]\). Khi đó: \(\mathop {\min }\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = \mathop {\min }\limits_{\left[ {2;\,\frac{5}{2}} \right]} h\left( t \right)\,\)

\(\mathop {max}\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = \mathop {max}\limits_{\left[ {2\,;\,\frac{5}{2}} \right]} h\left( t \right) = max\left\{ {\left| {m + 2} \right|\,;\,\left| {m + \frac{5}{2}} \right|} \right\} = \frac{{\left| {\left( {m + 2} \right) + \left( {m + \frac{5}{2}} \right)} \right| + \left| {\left( {m + 2} \right) - \left( {m + \frac{5}{2}} \right)} \right|}}{2} = \left| {m + \frac{9}{4}} \right| + \frac{1}{4}\)

*) Xét \(\left( {m + 2} \right)\left( {m + \frac{5}{2}} \right) \le 0 \Leftrightarrow m \in \left[ { - \frac{5}{2}\,;\, - 2} \right]\,\,\left( 1 \right)\)

Khi đó, \(\mathop {\min }\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = 0\). Suy ra

\(\mathop {\min }\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) + 2\mathop {max}\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = \frac{1}{2} \Leftrightarrow \left| {2m + \frac{9}{2}} \right| + \frac{1}{2} = \frac{1}{2} \Leftrightarrow m = - \frac{9}{4}\,\,\left( {thoa\,\,man\,\left( 1 \right)} \right)\)

*) Xét \(\left( {m + 2} \right)\left( {m + \frac{5}{2}} \right) > 0 \Leftrightarrow \left[ \begin{array}{l} m < - \frac{5}{2}\\ m > - 2 \end{array} \right.\,\,\,\left( 2 \right)\). Khi đó

\(\mathop {\min }\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = \mathop {\min }\limits_{\left[ {1\,;\,\frac{5}{2}} \right]} h\left( t \right) = \min \left\{ {\left| {m + 2} \right|\,;\,\left| {m + \frac{5}{2}} \right|} \right\} = \frac{{\left| {\left( {m + 2} \right) + \left( {m + \frac{5}{2}} \right)} \right| - \left| {\left( {m + 2} \right) - \left( {m + \frac{5}{2}} \right)} \right|}}{2} = \left| {m + \frac{9}{4}} \right| - \frac{1}{4}\)

Suy ra 

\(\begin{array}{l} \mathop {\min }\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) + 2\mathop {max}\limits_{\left[ {2\,;\,3} \right]} f\left( x \right) = \frac{1}{2}\\ \Leftrightarrow \left| {m + \frac{9}{4}} \right| - \frac{1}{4} + 2\left| {m + \frac{9}{4}} \right| + \frac{1}{2} = \frac{1}{2}\\ \Leftrightarrow \left| {m + \frac{9}{4}} \right| = \frac{1}{{12}}\\ \Leftrightarrow \left| {m + \frac{9}{4}} \right| = \frac{1}{{12}} \Leftrightarrow \left[ \begin{array}{l} m = - \frac{{13}}{6}\\ m = - \frac{7}{3} \end{array} \right.\,\left( L \right) \end{array}\)

Vậy \(S = \left\{ {\, - \frac{9}{4}\,} \right\}\). Suy ra, số phần tử của tập S bằng 1.