Đáp án:
ii. \(f.m \in \left( {\frac{{3 - \sqrt {39} }}{6};\frac{{3 + \sqrt {39} }}{6}} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
i.a.4{m^2} - \left( {m - 5} \right)\left( {m - 2} \right) \ge 0\\
\to 3{m^2} + 7m - 10 \ge 0\\
\to m \in \left( { - \infty ; - \frac{{10}}{3}} \right] \cup \left[ {1; + \infty } \right)\\
b.4 - 12m + 9{m^2} - \left( {m - 2} \right)\left( {5m - 6} \right) \ge 0\\
\to 4{m^2} + 4m - 8 \ge 0\\
\to m \in \left( { - \infty ; - 2} \right] \cup \left[ {1; + \infty } \right)\\
c.{m^2} + 6m + 9 - \left( {3 - m} \right)\left( {m + 2} \right) \ge 0\\
\to 2{m^2} + 5m + 3 \ge 0\\
\to m \in \left( { - \infty ; - \frac{3}{2}} \right] \cup \left[ { - 1; + \infty } \right)\\
d.{m^2} - \left( {1 + m} \right)\left( {2m} \right) \ge 0\\
\to - {m^2} - 2m \ge 0\\
\to m \in \left[ { - 2;0} \right]\\
e.4{m^2} - \left( {m - 2} \right)\left( {2m - 6} \right) \ge 0\\
\to 2{m^2} + 10m - 12 \ge 0\\
\to m \in \left( { - \infty ; - 6} \right] \cup \left[ {1; + \infty } \right)\\
f.4 - 12m + 9{m^2} + 3\left( { - {m^2} + 2m - 3} \right) \ge 0\\
\to 6{m^2} - 6m - 9 \ge 0\\
\to m \in \left( { - \infty ;\frac{{3 - \sqrt {39} }}{6}} \right] \cup \left[ {\frac{{3 + \sqrt {39} }}{6}; + \infty } \right)\\
ii.a.m \in \left( { - \frac{{10}}{3};1} \right)\\
b.m \in \left( { - 2;1} \right)\\
c.m \in \left( { - \frac{3}{2}; - 1} \right)\\
d.m \in \left( { - \infty ; - 2} \right) \cup \left( {0; + \infty } \right)\\
e.m \in \left( { - 6;1} \right)\\
f.m \in \left( {\frac{{3 - \sqrt {39} }}{6};\frac{{3 + \sqrt {39} }}{6}} \right)
\end{array}\)
