Đáp án:
$\begin{array}{l}
pt\,có\,2\,nghiệm\,pb\\
\Rightarrow \Delta ' = {\left( {m - 1} \right)^2} - 2{m^2} + 3m - 1 = - {m^2} + m > 0\\
\Rightarrow 0 < m < 1\\
theo\,viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_1}{x_2} = 2{m^2} - 3m + 1
\end{array} \right.\\
\Rightarrow P = \left| {2m - 2 + 2{m^2} - 3m + 1} \right| = \left| {2{m^2} - m - 1} \right| = \left| {2\left( {{m^2} - \frac{1}{2}m + \frac{1}{{16}}} \right) - \frac{9}{8}} \right|\\
= \left| {2{{\left( {m - \frac{1}{4}} \right)}^2} - \frac{9}{8}} \right|\\
do\,2{\left( {m - \frac{1}{4}} \right)^2} - \frac{9}{8} \ge - \frac{9}{8} \Rightarrow \left| {2{{\left( {m - \frac{1}{4}} \right)}^2} - \frac{9}{8}} \right| \ge \frac{9}{8}\\
dấu = \,xảy\,ra \Rightarrow m = \frac{1}{4}\left( {tm} \right)\\
Vậy\,\min P = \frac{9}{8}
\end{array}$
