Đáp án:

c) \( - 1 < m < 5\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:m \ne  \pm \dfrac{3}{2}\\
A = \dfrac{{{{\left( {2m - 3} \right)}^2} + {{\left( {2m + 3} \right)}^2}}}{{\left( {2m + 3} \right)\left( {2m - 3} \right)}}\\
 = \dfrac{{4{m^2} - 12m + 9 + 4{m^2} + 12m + 9}}{{\left( {2m + 3} \right)\left( {2m - 3} \right)}}\\
 = \dfrac{{8{m^2} + 18}}{{\left( {2m + 3} \right)\left( {2m - 3} \right)}}\\
A < 0\\
 \to \dfrac{{8{m^2} + 18}}{{\left( {2m + 3} \right)\left( {2m - 3} \right)}} < 0\\
 \to \left( {2m + 3} \right)\left( {2m - 3} \right) < 0\left( {8{m^2} + 18 > 0\forall m} \right)\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2m + 3 > 0\\
2m - 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2m + 3 < 0\\
2m - 3 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m >  - \dfrac{3}{2}\\
m < \dfrac{3}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
m <  - \dfrac{3}{2}\\
m > \dfrac{3}{2}
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to \dfrac{3}{2} > m >  - \dfrac{3}{2}\\
b)DK:m \ne \left\{ { - 8; - 3} \right\}\\
B = \dfrac{{\left( {1 - m} \right)\left( {m + 3} \right) + \left( {m - 1} \right)\left( {m + 8} \right)}}{{\left( {m + 8} \right)\left( {m + 3} \right)}}\\
 = \dfrac{{ - {m^2} - 2m + 3 + {m^2} + 7m - 8}}{{\left( {m + 8} \right)\left( {m + 3} \right)}}\\
 = \dfrac{{5m - 5}}{{\left( {m + 8} \right)\left( {m + 3} \right)}}\\
B > 0\\
 \to \dfrac{{5m - 5}}{{\left( {m + 8} \right)\left( {m + 3} \right)}} > 0\\
TH1:5m - 5 > 0 \to m > 1\\
 \to \left( {m + 8} \right)\left( {m + 3} \right) > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 8 > 0\\
m + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 8 < 0\\
m + 3 < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m >  - 8\\
m >  - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
m <  - 8\\
m <  - 3
\end{array} \right.
\end{array} \right.\\
 \to m > 1\\
TH2:5m - 5 < 0 \to m < 1\\
 \to \left( {m + 8} \right)\left( {m + 3} \right) < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 8 > 0\\
m + 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 8 < 0\\
m + 3 > 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
 - 3 > m >  - 8\\
\left\{ \begin{array}{l}
m <  - 8\\
m >  - 3
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to  - 3 > m >  - 8\\
KL:\left[ \begin{array}{l}
m > 1\\
 - 3 > m >  - 8
\end{array} \right.\\
c)\dfrac{{\left( {m + 1} \right)\left( {m - 5} \right)}}{2} < 0\\
 \to \left( {m + 1} \right)\left( {m - 5} \right) < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 1 > 0\\
m - 5 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 1 < 0\\
m - 5 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m >  - 1\\
m < 5
\end{array} \right.\\
\left\{ \begin{array}{l}
m <  - 1\\
m > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
 \to  - 1 < m < 5
\end{array}\)