Đáp án: $\,2 < m \le \dfrac{5}{2}$
Giải thích các bước giải:
Pt có 2 nghiệm phân biệt khi:
$\begin{array}{l}
\left\{ \begin{array}{l}
m - 1 \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
{\left( {m + 1} \right)^2} - \left( {m - 1} \right).3.\left( {m - 2} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
{m^2} + 2m + 1 - 3{m^2} + 9m - 6 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
2{m^2} - 11m + 5 < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
\left( {2m - 1} \right)\left( {m - 5} \right) < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 1\\
\dfrac{1}{2} < m < 5
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2\left( {m + 1} \right)}}{{m - 1}}\\
{x_1}{x_2} = \dfrac{{3\left( {m - 2} \right)}}{{m - 1}}
\end{array} \right.\\
\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} \ge \dfrac{{14}}{3}\\
\Rightarrow \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} - \dfrac{{14}}{3} \ge 0\\
\Rightarrow \dfrac{{2\left( {m + 1} \right)}}{{3\left( {m - 2} \right)}} - \dfrac{{14}}{3} \ge 0\\
\Rightarrow \dfrac{{m + 1 - 7\left( {m - 2} \right)}}{{m - 2}} \ge 0\\
\Rightarrow \dfrac{{m + 1 - 7m + 14}}{{m - 2}} \ge 0\\
\Rightarrow \dfrac{{6m - 15}}{{m - 2}} \le 0\\
\Rightarrow 2 < m \le \dfrac{5}{2}\\
Vậy\,2 < m \le \dfrac{5}{2}
\end{array}$
