Đáp án:
B10:
b) \( - \dfrac{3}{2} \le m \le - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
B10:\\
a)DK:\left\{ \begin{array}{l}
m \ne 3\\
{m^2} + 6m + 9 - \left( {3 - m} \right)\left( {m + 2} \right) > 0\\
\dfrac{{2m + 6}}{{3 - m}} > 0\\
\dfrac{{m + 2}}{{3 - m}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m \ne 3\\
{m^2} + 6m + 9 + {m^2} - m - 6 > 0\\
- 3 < m < 3\\
- 2 < m < 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2 < m < 3\\
2{m^2} + 5m + 3 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 2 < m < 3\\
\left[ \begin{array}{l}
m > - 1\\
m < - \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 1 < m < 3\\
- 2 < m < - \dfrac{3}{2}
\end{array} \right.\\
b)f\left( x \right) \le 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
3 - m < 0\\
{m^2} + 6m + 9 - \left( {3 - m} \right)\left( {m + 2} \right) \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3 < m\\
2{m^2} + 5m + 3 \le 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3 < m\\
- \dfrac{3}{2} \le m \le - 1
\end{array} \right.\\
\to - \dfrac{3}{2} \le m \le - 1\\
B11:\\
a)DK:2.\left( {3 + 4m + {m^2}} \right) < 0\\
\to 3 + 4m + {m^2} < 0\\
\to - 3 < m < - 1\\
b)DK:\left\{ \begin{array}{l}
{m^2} + 4m + 4 - 2\left( {3 + 4m + {m^2}} \right) \ge 0\\
m + 2 > 0\\
\dfrac{{{m^2} + 4m + 3}}{2} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- {m^2} - 4m - 2 \ge 0\\
m > - 2\\
\left[ \begin{array}{l}
m > - 1\\
m < - 3
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m \ge - 2 + \sqrt 2 \\
m \le - 2 - \sqrt 2
\end{array} \right.\\
m > - 2\\
\left[ \begin{array}{l}
m > - 1\\
m < - 3
\end{array} \right.
\end{array} \right.\\
\to m \ge - 2 + \sqrt 2
\end{array}\)
