Đáp án:
d) m=-26
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:\left( {m + 2} \right)\left( {m - 4} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m + 2 > 0\\
m - 4 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 2 < 0\\
m - 4 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > - 2\\
m < 4
\end{array} \right.\\
\left\{ \begin{array}{l}
m < - 2\\
m > 4
\end{array} \right.\left( l \right)
\end{array} \right.\\
b)DK:\left\{ \begin{array}{l}
\Delta ' > 0\\
m + 2 \ne 0\\
\frac{{2m + 2}}{{m + 2}} > 0\\
\frac{{m - 4}}{{m + 2}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 2m + 1 - \left( {m + 2} \right)\left( {m - 4} \right) > 0\\
m \ne - 2\\
\left[ \begin{array}{l}
m > - 1\\
m < - 2
\end{array} \right.\\
\left[ \begin{array}{l}
m > 4\\
m < - 2
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2} + 2m + 1 - {m^2} + 2m + 8 > 0\\
\left[ \begin{array}{l}
m > 4\\
m < - 2
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4m > - 9\\
\left[ \begin{array}{l}
m > 4\\
m < - 2
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m > - \frac{9}{4}\\
\left[ \begin{array}{l}
m > 4\\
m < - 2
\end{array} \right.
\end{array} \right.\\
d)\Delta ' > 0 \to m > - \frac{9}{4};m \ne - 2\\
3\left( {{x_1} + {x_2}} \right) = 5{x_1}{x_2}\\
\to 3\left( {\frac{{2m + 2}}{{m + 2}}} \right) = 5.\frac{{m - 4}}{{m + 2}}\\
\to 6m + 6 = 5m - 20\\
\to m = - 26
\end{array}\)
