Đáp án:

\[S = \left( { - \frac{2}{3};0} \right)\]

Giải thích các bước giải:

ĐKXĐ:  \( - {x^2} + 3x - 2 \ne 0 \Leftrightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 2
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
 - 1 < \frac{{10{x^2} - 3x - 2}}{{ - {x^2} + 3x - 2}} < 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
\frac{{10{x^2} - 3x - 2}}{{ - {x^2} + 3x - 2}} >  - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
\frac{{10{x^2} - 3x - 2}}{{ - {x^2} + 3x - 2}} < 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow \frac{{10{x^2} - 3x - 2}}{{ - {x^2} + 3x - 2}} + 1 > 0\\
 \Leftrightarrow \frac{{\left( {10{x^2} - 3x - 2} \right) + \left( { - {x^2} + 3x - 2} \right)}}{{ - {x^2} + 3x - 2}} > 0\\
 \Leftrightarrow \frac{{9{x^2} - 4}}{{ - {x^2} + 3x - 2}} > 0\\
 \Leftrightarrow \frac{{9{x^2} - 4}}{{{x^2} - 3x + 2}} < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
9{x^2} - 4 < 0\\
{x^2} - 3x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
9{x^2} - 4 > 0\\
{x^2} - 3x + 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
 - \frac{2}{3} < x < \frac{2}{3}\\
\left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \frac{2}{3}\\
x <  - \frac{2}{3}
\end{array} \right.\\
1 < x < 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
 - \frac{2}{3} < x < \frac{2}{3}\\
1 < x < 2
\end{array} \right.\\
 \Rightarrow {S_1} = \left( { - \frac{2}{3};\frac{2}{3}} \right) \cup \left( {1;2} \right)\\
\left( 2 \right) \Leftrightarrow \frac{{10{x^2} - 3x - 2}}{{ - {x^2} + 3x - 2}} - 1 < 0\\
 \Leftrightarrow \frac{{10{x^2} - 3x - 2 - \left( { - {x^2} + 3x - 2} \right)}}{{ - {x^2} + 3x - 2}} < 0\\
 \Leftrightarrow \frac{{11{x^2} - 6x}}{{ - {x^2} + 3x - 2}} < 0\\
 \Leftrightarrow \frac{{11{x^2} - 6x}}{{{x^2} - 3x + 2}} > 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
11{x^2} - 6x > 0\\
{x^2} - 3x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
11{x^2} - 6x < 0\\
{x^2} - 3x + 2 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x > \frac{6}{{11}}\\
x < 0
\end{array} \right.\\
\left[ \begin{array}{l}
x > 2\\
x < 1
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
0 < x < \frac{6}{{11}}\\
1 < x < 2
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x > 2\\
x < 0\\
\frac{6}{{11}} < x < 1
\end{array} \right.\\
 \Rightarrow {S_2} = \left( { - \infty ;0} \right) \cup \left( {\frac{6}{{11}};1} \right) \cup \left( {2; + \infty } \right)\\
S = {S_1} \cap {S_2} = \left( { - \frac{2}{3};0} \right)
\end{array}\)

Vậy tập nghiệm của bất phương trình đã cho là \(S = \left( { - \frac{2}{3};0} \right)\)