Đáp án:
a) \(\left[ \begin{array}{l}
2 < m < 3\\
1 < m < \dfrac{6}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
m \ne 2\\
4{m^2} - 12m + 9 - \left( {m - 2} \right)\left( {5m - 6} \right) > 0\\
\dfrac{{4m - 6}}{{m - 2}} > 0\\
\dfrac{{5m - 6}}{{m - 2}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
4{m^2} - 12m + 9 - 5{m^2} + 16m - 12 > 0\\
\left[ \begin{array}{l}
m > 2\\
m < \dfrac{3}{2}
\end{array} \right.\\
\left[ \begin{array}{l}
m > 2\\
m < \dfrac{6}{5}
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
- {m^2} + 4m - 3 > 0\\
\left[ \begin{array}{l}
m > 2\\
m < \dfrac{6}{5}
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
1 < m < 3\\
\left[ \begin{array}{l}
m > 2\\
m < \dfrac{6}{5}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
2 < m < 3\\
1 < m < \dfrac{6}{5}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
m \ne 2\\
\Delta ' \ge 0\\
{x_1}{x_2} > 0\\
{x_1} + {x_2} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
4{m^2} - 12m + 9 - \left( {m - 2} \right)\left( {5m - 6} \right) \ge 0\\
\dfrac{{5m - 6}}{{m - 2}} > 0\\
\dfrac{{4m - 6}}{{m - 2}} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 2\\
- {m^2} + 4m - 3 \ge 0\\
\left[ \begin{array}{l}
m > 2\\
m < \dfrac{6}{5}
\end{array} \right.\\
\dfrac{3}{2} < m < 2
\end{array} \right.\\
\to m \in \emptyset \\
c){x_1}^2{x_2} + {x_1}{x_2}^2 < 1\\
\to {x_1}{x_2}\left( {{x_1} + {x_2}} \right) < 1\\
\to \left( {\dfrac{{5m - 6}}{{m - 2}}} \right)\left( {\dfrac{{4m - 6}}{{m - 2}}} \right) < 1\\
\to \dfrac{{20{m^2} - 54m + 36 - {m^2} + 4m - 4}}{{{{\left( {m - 2} \right)}^2}}} < 0\\
\to \dfrac{{19{m^2} - 50m + 32}}{{{{\left( {m - 2} \right)}^2}}} < 0\\
\to 19{m^2} - 50m + 32 < 0\\
\left( {do:{{\left( {m - 2} \right)}^2} > 0\forall m \ne 2} \right)\\
\to \left[ \begin{array}{l}
m > \dfrac{{25 + \sqrt {17} }}{{19}}\\
m < \dfrac{{25 - \sqrt {17} }}{{19}}
\end{array} \right.
\end{array}\)
