Giải thích các bước giải:
\(\begin{array}{l}
a.\left[ \begin{array}{l}
\frac{{3x}}{{{x^2} - 4}} < 1\\
\frac{{3x}}{{{x^2} - 4}} > - 1
\end{array} \right. \to \left[ \begin{array}{l}
\frac{{3x - {x^2} + 4}}{{{x^2} - 4}} < 0\\
\frac{{3x + {x^2} - 4}}{{{x^2} - 4}} > 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{\left( {4 - x} \right)\left( {x + 1} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} < 0\left( 1 \right)\\
\frac{{\left( {x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} > 0\left( 2 \right)
\end{array} \right.
\end{array}\)
BXD:
x -∞ -2 -1 2 4 +∞
(1) - // + 0 - // + 0 -
\(KL:\left( 1 \right):x \in \left( { - \infty ; - 2} \right) \cup \left( { - 1;2} \right) \cup \left( {4; + \infty } \right)\)
BXD:
x -∞ -4 -2 1 2 +∞
(2) + // - 0 + // - 0 +
\(KL:\left( 2 \right):x \in \left( { - \infty ; - 4} \right) \cup \left( { - 2;1} \right) \cup \left( {2; + \infty } \right)\)
\(\begin{array}{l}
b.DK:x \ne \left\{ {2;3} \right\}\\
\left[ \begin{array}{l}
\frac{{x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} \ge 3\\
\frac{{x - 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} \le - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{1 - 3x + 9}}{{x - 3}} \ge 0\\
\frac{{1 + 3x - 9}}{{x - 3}} \le 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\frac{{10 - 3x}}{{x - 3}} \ge 0\\
\frac{{3x - 8}}{{x - 3}} \le 0
\end{array} \right. \to \left[ \begin{array}{l}
x \in \left[ {\frac{{10}}{3};3} \right)\\
x \in \left[ {\frac{8}{3};3} \right)
\end{array} \right.
\end{array}\)
