Đáp án: $m = \dfrac{3}{2}$
Giải thích các bước giải:
$\begin{array}{l}
b)Dk:m > - \dfrac{5}{8}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = 2m - 1
\end{array} \right.\\
Khi:\left| {{x_1}} \right| = 2\left| {{x_2}} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
{x_1} = 2{x_2}\\
{x_1} = - 2{x_2}
\end{array} \right.\\
+ TH1:{x_1} = 2{x_2}\\
\Leftrightarrow \left\{ \begin{array}{l}
2{x_2} + {x_2} = 3\\
{x_1}{x_2} = 2m - 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = 1;{x_1} = 2\\
1.2 = 2m - 1
\end{array} \right.\\
\Leftrightarrow 2m = 3\\
\Leftrightarrow m = \dfrac{3}{2}\left( {tmdk} \right)\\
+ TH2:{x_1} = - 2{x_2}\\
\Leftrightarrow - 2{x_2} + {x_2} = 3\\
\Leftrightarrow {x_2} = - 3\\
\Leftrightarrow {x_1} = 6\\
\Leftrightarrow 6.\left( { - 3} \right) = 2m - 1\\
\Leftrightarrow 2m = - 17\\
\Leftrightarrow m = \dfrac{{ - 17}}{2}\left( {ktm} \right)\\
Vậy\,m = \dfrac{3}{2}
\end{array}$
