Đáp án:
e) \(\dfrac{{13}}{{12}} \ge m > 1\)
Giải thích các bước giải:
\(\begin{array}{l}
d)DK:\left\{ \begin{array}{l}
m \ne 3\\
{m^2} + 6m + 9 - \left( {3 - m} \right)\left( {m + 2} \right) > 0\\
\dfrac{{2m + 6}}{{3 - m}} > 0\\
\dfrac{{m + 2}}{{3 - m}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 3\\
{m^2} + 6m + 9 + {m^2} - m - 6 > 0\\
- 3 < m < 3\\
- 2 < m < 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2 < m < 3\\
2{m^2} + 5m + 3 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2 < m < 3\\
\left( {m + 1} \right)\left( {2m + 3} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2 < m < 3\\
\left[ \begin{array}{l}
m > - 1\\
m < - \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 1 < m < 3\\
- 2 < m < - \dfrac{3}{2}
\end{array} \right.\\
b)DK:\left\{ \begin{array}{l}
m \ne 1\\
4{m^2} - 12m + 9 - 4\left( {m - 1} \right)\left( {m + 1} \right) \ge 0\\
\dfrac{{ - 2m + 3}}{{m - 1}} > 0\\
\dfrac{{m + 1}}{{m - 1}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 1\\
4{m^2} - 12m + 9 - 4{m^2} + 4 \ge 0\\
1 < m < \dfrac{3}{2}\\
\left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 < m < \dfrac{3}{2}\\
- 12m + 13 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
1 < m < \dfrac{3}{2}\\
\dfrac{{13}}{{12}} \ge m
\end{array} \right.\\
\to \dfrac{{13}}{{12}} \ge m > 1
\end{array}\)
