Đáp án:
\(2 \le m \le \dfrac{5}{2}\)
Giải thích các bước giải:
Để phương trình có 2 nghiệm phân biệt
⇔Δ'>0 và \(m \ne 1\)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
{m^2} + 2m + 1 - \left( {m - 1} \right).3\left( {m - 2} \right) > 0\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 2m + 1 - 3{m^2} + 9m - 6 > 0\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2{m^2} + 11m - 5 > 0\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {5 - m} \right)\left( {2m - 1} \right) > 0\\
m \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( {\dfrac{1}{2};5} \right)\\
m \ne 1
\end{array} \right.\\
Có:\dfrac{1}{{{x_1}}} + \dfrac{1}{{{x_2}}} \ge \dfrac{{14}}{3}\\
\to \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} \ge \dfrac{{14}}{3}\\
\to \dfrac{{\dfrac{{2m + 2}}{{m - 1}}}}{{\dfrac{{3m - 6}}{{m - 1}}}} \ge \dfrac{{14}}{3}\\
\to \dfrac{{2m + 2}}{{3m - 6}} \ge \dfrac{{14}}{3}\\
\to \dfrac{{6m + 6 - 42m + 84}}{{3m - 6}} \ge 0\\
\to \dfrac{{ - 36m + 90}}{{3m - 6}} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 36m + 90 \ge 0\\
3m - 6 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- 36m + 90 \le 0\\
3m - 6 \le 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \le \dfrac{5}{2}\\
m \ge 2
\end{array} \right.\\
\left\{ \begin{array}{l}
m \ge \dfrac{5}{2}\\
m \le 2
\end{array} \right.\left( l \right)
\end{array} \right.\\
KL:2 \le m \le \dfrac{5}{2}
\end{array}\)
