Đáp án:
4) \(1 \le x < 9\)
Giải thích các bước giải:
\(\begin{array}{l}
2)DK:x \ne \left\{ { - 2;3} \right\}\\
\dfrac{{{x^2} - x - 3 - {x^2} + x + 6}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} \le 0\\
\to \dfrac{3}{{\left( {x + 2} \right)\left( {x - 3} \right)}} \le 0\\
\to \left( {x + 2} \right)\left( {x - 3} \right) < 0\\
\to x \in \left( { - 2;3} \right)\\
3)DK:x \ne \left\{ {3;4} \right\}\\
\dfrac{{x - 4 + x - 3}}{{\left( {x - 3} \right)\left( {x - 4} \right)}} \le 0\\
\to \dfrac{{2x - 7}}{{\left( {x - 3} \right)\left( {x - 4} \right)}} \le 0
\end{array}\)
BXD
x -∞ 3 7/2 4 +∞
f(x) - // + 0 - // +
\(\begin{array}{l}
\to x \in \left( { - \infty ;3} \right) \cup \left[ {\dfrac{7}{2};4} \right)\\
4)\left\{ \begin{array}{l}
\dfrac{{2x + 1 - 3}}{3} \ge 0\\
\dfrac{{x + 5 - 14}}{2} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x - 2 \ge 0\\
x - 9 < 0
\end{array} \right.\\
\to 1 \le x < 9
\end{array}\)
