Đáp án:
\(\left[ \begin{array}{l}
m > 1\\
- \dfrac{1}{3} \le m < 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\Delta ' \ge 0\\
\to \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 2m + 1 - m\left( {m - 5} \right) \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 2m + 1 - {m^2} + 5m \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
3m + 1 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
m \ge - \dfrac{1}{3}
\end{array} \right.\\
Có:{x_1} < {x_2} < 2\\
\to \left\{ \begin{array}{l}
{x_1} - 2 < 0\\
{x_2} - 2 < 0
\end{array} \right.\\
\to \left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) > 0\\
\to {x_1}{x_2} - 2\left( {{x_1} + {x_1}} \right) + 4 > 0\\
\to \dfrac{{m - 5}}{m} - 2.\dfrac{{ - 2m + 2}}{m} + 4 > 0\\
\to \dfrac{{m - 5 + 4m - 4 + 4m}}{m} > 0\\
\to \dfrac{{9m - 9}}{m} > 0\\
\to \left[ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.\\
KL:\left[ \begin{array}{l}
m > 1\\
- \dfrac{1}{3} \le m < 0
\end{array} \right.
\end{array}\)
