Giải thích các bước giải:

Ta có:

1,

ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne  - 1\\
x \ne \dfrac{5}{2}
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
\dfrac{{2x - 5}}{{x + 1}} \ge \dfrac{{x + 1}}{{2x - 5}}\\
 \Leftrightarrow \dfrac{{2x - 5}}{{x + 1}} - \dfrac{{x + 1}}{{2x - 5}} \ge 0\\
 \Leftrightarrow \dfrac{{{{\left( {2x - 5} \right)}^2} - {{\left( {x + 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {2x - 5} \right)}} \ge 0\\
 \Leftrightarrow \dfrac{{\left[ {\left( {2x - 5} \right) - \left( {x + 1} \right)} \right].\left[ {\left( {2x - 5} \right) + \left( {x + 1} \right)} \right]}}{{\left( {x + 1} \right)\left( {2x - 5} \right)}} \ge 0\\
 \Leftrightarrow \dfrac{{\left( {x - 6} \right).\left( {3x - 4} \right)}}{{\left( {x + 1} \right)\left( {2x - 5} \right)}} \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left( {x - 6} \right)\left( {3x - 4} \right) \ge 0\\
\left( {x + 1} \right)\left( {2x - 5} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\left( {x - 6} \right)\left( {3x - 4} \right) \le 0\\
\left( {x + 1} \right)\left( {2x - 5} \right) < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 6\\
x \le \dfrac{4}{3}
\end{array} \right.\\
\left[ \begin{array}{l}
x > \dfrac{5}{2}\\
x <  - 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\dfrac{4}{3} \le x \le 6\\
 - 1 < x < \dfrac{5}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \ge 6\\
x <  - 1\\
\dfrac{4}{3} \le x < \dfrac{5}{2}
\end{array} \right.\\
 \Rightarrow S = \left( { - \infty ; - 1} \right) \cup \left[ {\dfrac{4}{3};\dfrac{5}{2}} \right) \cup \left[ {6; + \infty } \right)
\end{array}\)

2,

Ta có:

\(\begin{array}{l}
f\left( x \right) = a\,{x^2} + bx + c < 0,\,\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}
a < 0\\
\Delta  < 0
\end{array} \right.\\
\left( {m - 3} \right){x^2} - 2mx + m - 6 < 0,\,\,\,\forall x \in R\\
 \Leftrightarrow \left\{ \begin{array}{l}
m - 3 < 0\\
\Delta ' < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < 3\\
{m^2} - \left( {m - 3} \right)\left( {m - 6} \right) < 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m < 3\\
{m^2} - \left( {{m^2} - 9m + 18} \right) < 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m < 3\\
9m - 18 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m < 3\\
m < 2
\end{array} \right. \Leftrightarrow m < 2
\end{array}\)