Giải thích các bước giải:
$\begin{array}{l} \left\{ \begin{array}{l} \frac{{x + 1}}{{x + 2}} \ge \frac{{2x + 1}}{{2x - 1}}\\ \frac{{2x - 1}}{{x + 3}} < 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \frac{{x + 1}}{{x + 2}} - \frac{{2x + 1}}{{2x - 1}} \ge 0\\ \frac{{2x - 1}}{{2x + 6}} < 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \frac{{(x + 1)(2x - 1) - (2x + 1)(x + 2)}}{{(x + 2)(2x - 1)}} \ge 0\\ \left\{ \begin{array}{l} 2x - 1 < 0\\ 2x + 6 > 0 \end{array} \right.(do\,2x - 1 < 2x + 6) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \frac{{(2{x^2} + x - 1) - (2{x^2} + 5x + 2)}}{{(x + 2)(2x - 1)}} \ge 0\\ \left\{ \begin{array}{l} x < \frac{1}{2}\\ x > - 3 \end{array} \right. \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \frac{{ - 4x - 3}}{{(x + 2)(2x - 1)}} \ge 0\\ - 3 < x < \frac{1}{2} \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x < - 2\\ - \frac{3}{4} \le x < \frac{1}{2} \end{array} \right.\\ - 3 < x < \frac{1}{2} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} - 3 < x < - 2\\ - \frac{3}{4} \le x < \frac{1}{2} \end{array} \right. \end{array}$
