Đáp án: $\dfrac{{ - 14}}{3} < m < 2 - \sqrt {41} $
Giải thích các bước giải:
Pt có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\Delta ' > 0\\
\Leftrightarrow {\left( {m + 5} \right)^2} - {m^2} - 4m + 3 > 0\\
\Leftrightarrow {m^2} + 10m + 25 - {m^2} - 4m + 3 > 0\\
\Leftrightarrow 6m + 28 > 0\\
\Leftrightarrow m > \dfrac{{ - 14}}{3}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 10\\
{x_1}{x_2} = {m^2} + 4m - 3
\end{array} \right.\\
Khi: - 2 < {x_1} < {x_2} < 4\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4 < {x_1} + {x_2} < 8\\
\left( {{x_1} + 2} \right)\left( {{x_2} + 2} \right) > 0\\
\left( {{x_1} - 4} \right)\left( {{x_2} - 4} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 4 < 2m + 10 < 8\\
{x_1}{x_2} + 2\left( {{x_1} + {x_2}} \right) + 4 > 0\\
{x_1}{x_2} - 4\left( {{x_1} + {x_2}} \right) + 16 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 7 < m < - 1\\
{m^2} + 4m - 3 + 4m + 20 + 4 > 0\\
{m^2} + 4m - 3 - 8m - 40 + 16 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 7 < m < - 1\\
{m^2} + 8m + 21 > 0\left( {tm} \right)\\
{m^2} - 4m - 37 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 7 < m < - 1\\
{\left( {m - 2} \right)^2} > 41
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 7 < m < - 1\\
\left[ \begin{array}{l}
m > 2 + \sqrt {41} \\
m < 2 - \sqrt {41}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow - 7 < m < 2 - \sqrt {41} \\
Vậy\,\dfrac{{ - 14}}{3} < m < 2 - \sqrt {41}
\end{array}$
