Đáp án:
\[S = \left( { - \infty ; - 7} \right) \cup \left( { - \frac{2}{3};\frac{3}{5}} \right) \cup \left( {\frac{5}{2}; + \infty } \right)\]
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{2x - 5}}{{3x + 2}} < \frac{{3x + 2}}{{2x - 5}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ne \frac{{ - 2}}{3};\,\,x \ne \frac{5}{2}} \right)\\
\Leftrightarrow \frac{{2x - 5}}{{3x + 2}} - \frac{{3x + 2}}{{2x - 5}} < 0\\
\Leftrightarrow \frac{{{{\left( {2x - 5} \right)}^2} - {{\left( {3x + 2} \right)}^2}}}{{\left( {3x + 2} \right)\left( {2x - 5} \right)}} < 0\\
\Leftrightarrow \frac{{\left[ {\left( {2x - 5} \right) - \left( {3x + 2} \right)} \right]\left[ {\left( {2x - 5} \right) + \left( {3x + 2} \right)} \right]}}{{\left( {3x + 2} \right)\left( {2x - 5} \right)}} < 0\\
\Leftrightarrow \frac{{\left( { - x - 7} \right)\left( {5x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x - 5} \right)}} < 0\\
\Leftrightarrow \frac{{\left( {x + 7} \right)\left( {5x - 3} \right)}}{{\left( {3x + 2} \right)\left( {2x - 5} \right)}} > 0\\
TH1:\,\,\,\left\{ \begin{array}{l}
\left( {x + 7} \right)\left( {5x - 3} \right) > 0\\
\left( {3x + 2} \right)\left( {2x - 5} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \frac{3}{5}\\
x < - 7
\end{array} \right.\\
\left[ \begin{array}{l}
x > \frac{5}{2}\\
x < - \frac{2}{3}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > \frac{5}{2}\\
x < - 7
\end{array} \right.\\
TH2:\,\,\,\,\left\{ \begin{array}{l}
\left( {x + 7} \right)\left( {5x - 3} \right) < 0\\
\left( {3x + 2} \right)\left( {2x - 5} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 7 < x < \frac{3}{5}\\
- \frac{2}{3} < x < \frac{5}{2}
\end{array} \right. \Leftrightarrow - \frac{2}{3} < x < \frac{3}{5}\\
\Rightarrow S = \left( { - \infty ; - 7} \right) \cup \left( { - \frac{2}{3};\frac{3}{5}} \right) \cup \left( {\frac{5}{2}; + \infty } \right)
\end{array}\)
