Đáp án:
1) \(m \in \left( {1;2} \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\left\{ \begin{array}{l}
2 - m > 0\\
{m^2} - 4m + 4 - \left( {2 - m} \right).m < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 2\\
{m^2} - 4m + 4 - 2m + {m^2} < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 2\\
2{m^2} - 6m + 4 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 2\\
2\left( {m - 2} \right)\left( {m - 1} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m < 2\\
m \in \left( {1;2} \right)
\end{array} \right.\\
\to m \in \left( {1;2} \right)\\
2)DK:\left\{ \begin{array}{l}
2m + 3 > 0\\
4{m^2} + 12m + 9 - \left( {2m + 3} \right)\left( {m + 1} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{3}{2}\\
4{m^2} + 12m + 9 - 2{m^2} - 5m - 3 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{3}{2}\\
2{m^2} + 7m + 6 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{3}{2}\\
\left( {2m + 3} \right)\left( {m + 2} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \dfrac{3}{2}\\
m \in \left[ { - 2; - \dfrac{3}{2}} \right]
\end{array} \right.\left( {vô lý} \right)
\end{array}\)
⇒ Không tồn tại m để bpt vô nghiệm
