Đáp án:
\(x \in \left( { - \infty ; - \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { - \frac{1}{3};\frac{1}{2}} \right\}\\
\frac{{x + 2}}{{3x + 1}} < \frac{{x - 2}}{{2x - 1}}\\
\to \frac{{\left( {x + 2} \right)\left( {2x - 1} \right) - \left( {x - 2} \right)\left( {3x + 1} \right)}}{{\left( {3x + 1} \right)\left( {2x - 1} \right)}} < 0\\
\to \frac{{2{x^2} + 3x - 2 - 3{x^2} + 5x + 2}}{{\left( {3x + 1} \right)\left( {2x - 1} \right)}} < 0\\
\to \frac{{ - {x^2} + 8x}}{{\left( {3x + 1} \right)\left( {2x - 1} \right)}} < 0\\
Xét: - {x^2} + 8x = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 8
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1/3 0 1/2 8 +∞
f(x) - // + 0 - // + 0 -
\(KL:x \in \left( { - \infty ; - \frac{1}{3}} \right) \cup \left( {0;\frac{1}{2}} \right) \cup \left( {8; + \infty } \right)\)
