Đáp án:
e) \(\dfrac{1}{3} \ge m > 0\)
Giải thích các bước giải:
\(\begin{array}{l}
d)DK;\left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 2m + 1 - m\left( { - m + 1} \right) > 0\\
\dfrac{{2m - 2}}{m} > 0\\
\dfrac{{1 - m}}{m} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
{m^2} - 2m + 1 + {m^2} - m > 0\\
\left[ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.\\
0 < m < 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 0\\
- 3m + 1 > 0\\
\left[ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.\\
0 < m < 1
\end{array} \right.\left( {vô lý} \right)\\
\to m \in \emptyset \\
e)Xét:\Delta ' \ge 0\\
\to - 3m + 1 \ge 0\\
\to \dfrac{1}{3} \ge m\\
2\left( {{x_1} + {x_2}} \right) + {x_1}{x_2} < 3\\
\to 2\left( {\dfrac{{2m - 2}}{m}} \right) + \dfrac{{1 - m}}{m} < 3\\
\to \dfrac{{4m - 4 + 1 - m - 3m}}{m} < 0\\
\to \dfrac{{ - 3}}{m} < 0\\
\to m > 0\\
KL:\dfrac{1}{3} \ge m > 0
\end{array}\)
