Đáp án: $m \in \left\{ { - \sqrt 2 ; - 1;1;\sqrt 2 } \right\}$
Giải thích các bước giải:
$\begin{array}{l}
b)2{x^2} - 3x - {m^2} + 1 = 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{2}\\
{x_1}{x_2} = \dfrac{{ - {m^2} + 1}}{2}
\end{array} \right.\\
+ Khi:{x_1} \le 0 < {x_2}\\
\left| {{x_1}} \right| + 2\left| {{x_2}} \right| = 3\\
\Leftrightarrow - {x_1} + 2{x_2} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
- {x_1} + 2{x_2} = 3\\
{x_1} + {x_2} = \dfrac{3}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3{x_2} = \dfrac{9}{2}\\
{x_1} = \dfrac{3}{2} - {x_2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{3}{2}\\
{x_1} = 0
\end{array} \right.\left( {tm} \right)\\
\Leftrightarrow {x_1}.{x_2} = - {m^2} + 1 = 0\\
\Leftrightarrow m = \pm 1\\
+ Khi:{x_2} \le 0 < {x_1}\\
\left| {{x_1}} \right| + 2\left| {{x_2}} \right| = 3\\
\Leftrightarrow {x_1} - 2{x_2} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} - 2{x_2} = 3\\
{x_1} + {x_2} = \dfrac{3}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 3{x_2} = \dfrac{3}{2}\\
{x_1} = \dfrac{3}{2} - {x_2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = - \dfrac{1}{2}\left( {tm} \right)\\
{x_1} = 2
\end{array} \right.\\
\Leftrightarrow 2.\dfrac{{ - 1}}{2} = - {m^2} + 1\\
\Leftrightarrow {m^2} = 2\\
\Leftrightarrow m = \pm \sqrt 2 \\
+ Khi:0 < {x_1};{x_2}\\
\Leftrightarrow {x_1} + 2{x_2} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{3}{2}\\
{x_1} + 2{x_2} = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_2} = \dfrac{3}{2}\\
{x_1} = 0
\end{array} \right.\left( {ktm} \right)\\
Vậy\,m \in \left\{ { - \sqrt 2 ; - 1;1;\sqrt 2 } \right\}
\end{array}$
