Đáp án:
d. \(m \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {m - 2} \right)\left( {m - 5} \right) < 0\\
\to m \in \left( {2;5} \right)\\
b.\left\{ \begin{array}{l}
4{m^2} - \left( {m - 5} \right)\left( {m - 2} \right) > 0\\
m \ne 5\\
\frac{{4m}}{{m - 5}} < 0\\
\frac{{m - 2}}{{m - 5}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 5\\
3{m^2} + 7m - 10 > 0\\
m \in \left( {0;5} \right)\\
m \in \left( { - \infty ;2} \right) \cup \left( {5; + \infty } \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 1} \right)\left( {3m + 10} \right) > 0\\
m \in \left( {0;2} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \in \left( { - \infty ; - \frac{{10}}{3}} \right) \cup \left( {1; + \infty } \right)\\
m \in \left( {0;2} \right)
\end{array} \right.\\
KL:m \in \left( {1;2} \right)\\
c.4{m^2} - \left( {m - 5} \right)\left( {m - 2} \right) \ge 0\\
\to 3{m^2} + 7m - 10 \ge 0\\
\to \left( {m - 1} \right)\left( {3m + 10} \right) \ge 0\\
\to m \in \left( { - \infty ; - \frac{{10}}{3}} \right] \cup \left[ {1; + \infty } \right)\\
d.\left\{ \begin{array}{l}
m - 5 > 0\\
3{m^2} + 7m - 10 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 5\\
\left( {m - 1} \right)\left( {3m + 10} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 5\\
m \in \left[ { - \frac{{10}}{3};1} \right]
\end{array} \right.\\
\to m \in \emptyset
\end{array}\)
