Giải thích các bước giải:
\(\begin{array}{l}
TH1:\,\,\,x < 1 \Rightarrow 1 - x > 0 \Rightarrow \left| {1 - x} \right| = 1 - x\\
\frac{x}{{\left| {1 - x} \right| + 2}} < 1\\
\Leftrightarrow \frac{x}{{1 - x + 2}} < 1\\
\Leftrightarrow \frac{x}{{3 - x}} - 1 < 0\\
\Leftrightarrow \frac{{x - \left( {3 - x} \right)}}{{3 - x}} < 0\\
\Leftrightarrow \frac{{2x - 3}}{{x - 3}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 3\\
x < \frac{3}{2}
\end{array} \right.\\
\Rightarrow {S_1} = \left( { - \infty ;\frac{3}{2}} \right) \cup \left( {3; + \infty } \right)\\
TH2:\,\,\,x \ge 1 \Rightarrow 1 - x \le 0 \Rightarrow \left| {1 - x} \right| = x - 1\\
\frac{x}{{\left| {1 - x} \right| + 2}} < 1\\
\Leftrightarrow \frac{x}{{x - 1 + 2}} < 1\\
\Leftrightarrow \frac{x}{{x + 1}} - 1 < 0\\
\Leftrightarrow \frac{{ - 1}}{{x + 1}} < 0\\
\Leftrightarrow x + 1 > 0\\
\Leftrightarrow x > - 1\\
\Rightarrow {S_2} = \left( { - 1; + \infty } \right)\\
S = {S_1} \cup {S_2} = R
\end{array}\)
