Đáp án:
c) m>2
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:1.\left( {2m - 4} \right) < 0\\
\to 2m - 4 < 0\\
\to m < 2\\
b)DK:\left\{ \begin{array}{l}
\Delta \ge 0\\
2m + 1 < 0\\
2m - 4 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4{m^2} + 4m + 1 - 4\left( {2m - 4} \right) \ge 0\\
m < - \dfrac{1}{2}\\
m > 2
\end{array} \right.\left( {vô lý} \right)\\
\to m \in \emptyset \\
c)DK:\left\{ \begin{array}{l}
\Delta \ge 0\\
2m + 1 > 0\\
2m - 4 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4{m^2} + 4m + 1 - 4\left( {2m - 4} \right) \ge 0\\
m > - \dfrac{1}{2}\\
m > 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 2\\
4{m^2} + 4m + 1 - 8m + 16 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 2\\
4{m^2} - 4m + 17 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 2\\
4{m^2} - 4m + 1 + 16 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > 2\\
{\left( {2m - 1} \right)^2} + 16 \ge 0\left( {ld} \right)\forall m \in R
\end{array} \right.\\
\to m > 2
\end{array}\)
