Đáp án:
$\begin{array}{l}
2{x^2} + 4x - m = 0\\
\Delta ' > 0\\
\Rightarrow {2^2} - 2.\left( { - m} \right) > 0\\
\Rightarrow 4 + 2m > 0\\
\Rightarrow m > - 2\\
TheoViet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_1}{x_2} = \dfrac{{ - m}}{2}
\end{array} \right.\\
a)3{x_1} + {x_2} = 0\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
3{x_1} + {x_2} = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2{x_1} = 2\\
{x_2} = - 3{x_1}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} = 1\\
{x_2} = - 3
\end{array} \right.\\
\Rightarrow {x_1}.{x_2} = - \dfrac{m}{2} = - 3\\
\Rightarrow m = 6\left( {tmdk} \right)\\
Vậy\,m = 6\\
b){x_2} = 3{x_1}\\
\Rightarrow {x_1} + 3{x_1} = - 2\\
\Rightarrow 4{x_1} = - 2\\
\Rightarrow {x_1} = - \dfrac{1}{2}\\
\Rightarrow {x_2} = 3{x_1} = - \dfrac{3}{2}\\
\Rightarrow \left( { - \dfrac{1}{2}} \right).\left( { - \dfrac{3}{2}} \right) = - \dfrac{m}{2}\\
\Rightarrow m = - \dfrac{3}{2}\left( {tmdk} \right)\\
Vậy\,m = - \dfrac{3}{2}\\
c){x_2} - {x_1} = 8\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\\
{x_2} - {x_1} = 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2{x_2} = 6\\
{x_1} = - 2 - {x_2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_2} = 3\\
{x_1} = - 5
\end{array} \right.\\
\Rightarrow \left( { - 5} \right).3 = - \dfrac{m}{2}\\
\Rightarrow m = 30\left( {tmdk} \right)\\
Vậy\,m = 30
\end{array}$
