Đáp án:
$\begin{array}{l}
\left( {m + 1} \right).{x^2} - 2\left( {m - 1} \right).x + m - 3 = 0\\
\Rightarrow \left\{ \begin{array}{l}
m + 1 \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{\left( {m - 1} \right)^2} - \left( {m + 1} \right)\left( {m - 3} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{m^2} - 2m + 1 - {m^2} + 2m + 3 > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne - 1\\
4 > 0\left( {tm} \right)
\end{array} \right.\\
\Rightarrow m \ne - 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2\left( {m - 1} \right)}}{{m + 1}}\\
{x_1}{x_2} = \dfrac{{m - 3}}{{m + 1}}
\end{array} \right.\\
DO:\left\{ \begin{array}{l}
{x_1}.{x_2} < 0\\
{x_1} = 2{x_2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
3{x_2} = \dfrac{{2\left( {m - 1} \right)}}{{m + 1}}\\
\dfrac{{m - 3}}{{m + 1}} < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{{2\left( {m - 1} \right)}}{{3\left( {m + 1} \right)}}\\
- 1 < m < 3\\
{x_1}.{x_2} = \dfrac{{m - 3}}{{m + 1}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 3\\
2.\dfrac{{2\left( {m - 1} \right)}}{{3\left( {m + 1} \right)}}.\dfrac{{2\left( {m - 1} \right)}}{{3\left( {m + 1} \right)}} = \dfrac{{m - 3}}{{m + 1}}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 3\\
\dfrac{8}{9}.\dfrac{{{{\left( {m - 1} \right)}^2}}}{{m + 1}} = m - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 3\\
8\left( {{m^2} - 2m + 1} \right) = 9\left( {{m^2} - 2m - 3} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 3\\
{m^2} - 2m - 35 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 3\\
m = 7/m = - 5
\end{array} \right.\left( {ktm} \right)
\end{array}$
Vậy ko có giá trị của m để thỏa mãn yêu cầu
