Đáp án:
\[S = \left( { - \infty ; - 2} \right) \cup \left( { - \frac{1}{2}; + \infty } \right)\]
Giải thích các bước giải:
ĐKXĐ: \(x \ne - 2\)
Ta có:
\(\begin{array}{l}
\frac{{\left| {x - 1} \right|}}{{x + 2}} < 1\\
\Leftrightarrow \frac{{\left| {x - 1} \right| - \left( {x + 2} \right)}}{{x + 2}} < 0\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,\,x < - 2 \Rightarrow \left\{ \begin{array}{l}
x + 2 < 0\\
x - 1 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left| {x - 1} \right| - \left( {x + 2} \right) > 0\\
\Leftrightarrow \left( {1 - x} \right) - \left( {x + 2} \right) > 0\\
\Leftrightarrow - 1 - 2x > 0\\
\Leftrightarrow 1 + 2x < 0\\
\Leftrightarrow x < - \frac{1}{2}\\
\Rightarrow x < - 2\\
TH2:\,\,\,\,\, - 2 < x < 1\,\,\,\,\, \Rightarrow \left\{ \begin{array}{l}
x + 2 > 0\\
x - 1 < 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left| {x - 1} \right| - \left( {x + 2} \right) < 0\\
\Leftrightarrow \left( {1 - x} \right) - \left( {x + 2} \right) < 0\\
\Leftrightarrow - 1 - 2x < 0\\
\Leftrightarrow 1 + 2x > 0\\
\Leftrightarrow x > - \frac{1}{2}\\
\Rightarrow - \frac{1}{2} < x < 1\\
TH3:\,\,\,\,\,x \ge 1 \Rightarrow \left\{ \begin{array}{l}
x + 2 > 0\\
x - 1 \ge 0
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \left| {x - 1} \right| - \left( {x + 2} \right) < 0\\
\Leftrightarrow x - 1 - x - 2 < 0\\
\Leftrightarrow - 3 < 0,\,\,\,\,\,t/m,\forall x \ge 1
\end{array}\)
Vậy tập nghiệm của BPT đã cho là \(S = \left( { - \infty ; - 2} \right) \cup \left( { - \frac{1}{2}; + \infty } \right)\)
