Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
m \ne 0\\
\Delta ' > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{\left( {m + 1} \right)^2} - m.m.{\left( {m + 1} \right)^2} > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
{\left( {m + 1} \right)^2}.\left( {1 - {m^2}} \right) > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
m \ne - 1\\
1 - {m^2} > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0;m \ne - 1\\
{m^2} < 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 1 < m < 1\\
m \ne 0
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2\left( {m + 1} \right)}}{m}\\
{x_1}{x_2} = \dfrac{{m{{\left( {m + 1} \right)}^2}}}{m} = {\left( {m + 1} \right)^2}
\end{array} \right.\\
{x_1} = 3{x_2}\\
\Rightarrow 4{x_2} = \dfrac{{2\left( {m + 1} \right)}}{m}\\
\Rightarrow {x_2} = \dfrac{{m + 1}}{{2m}}\\
\Rightarrow {x_1} = \dfrac{{3\left( {m + 1} \right)}}{{2m}}\\
\Rightarrow \dfrac{{m + 1}}{{2m}}.\dfrac{{3\left( {m + 1} \right)}}{{2m}} = {\left( {m + 1} \right)^2}\\
\Rightarrow {\left( {m + 1} \right)^2}.\left( {\dfrac{3}{{4{m^2}}} - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - 1\left( {ktm} \right)\\
{m^2} = \dfrac{4}{3}\left( {ktm} \right)
\end{array} \right.
\end{array}$
Vậy không có giá trị của m để thỏa mãn điều kiện.
