Đáp án:
$\begin{array}{l}
a)Khi:m = 0\\
\Rightarrow - 2x + 1 = 0\\
\Rightarrow x = \dfrac{1}{2}\left( {tm} \right)\\
+ Khi:m \ne 0\\
\Rightarrow \Delta ' = {\left( {m + 1} \right)^2} - m\left( {1 - 3m} \right)\\
= {m^2} + 2m + 1 - m + 3{m^2}\\
= 4{m^2} + m + 1\\
= 4{m^2} + 2.2m.\dfrac{1}{4} + \dfrac{1}{{16}} + \dfrac{{15}}{{16}}\\
= {\left( {2m + \dfrac{1}{4}} \right)^2} + \dfrac{{15}}{{16}} > 0
\end{array}$
Vậy phương trình luôn có nghiệm với mọi m
b)
Phương trình có 2 nghiệm phân biệt khi: m khác 0
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{2\left( {m + 1} \right)}}{m}\\
{x_1}{x_2} = \dfrac{{1 - 3m}}{m}
\end{array} \right.\\
A = x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= \dfrac{{4{{\left( {m + 1} \right)}^2}}}{{{m^2}}} - 2.\dfrac{{1 - 3m}}{m}\\
= \dfrac{{4{m^2} + 8m + 4}}{{{m^2}}} - \dfrac{2}{m} + 6\\
= 4 + \dfrac{8}{m} + \dfrac{4}{{{m^2}}} - \dfrac{2}{m} + 6\\
= \dfrac{4}{{{m^2}}} + \dfrac{6}{m} + 10\\
= {\left( {\dfrac{2}{m}} \right)^2} + 2.\dfrac{2}{m}.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{31}}{4}\\
= {\left( {\dfrac{2}{m} + \dfrac{3}{2}} \right)^2} + \dfrac{{31}}{4} \ge \dfrac{{31}}{4}\\
\Rightarrow \min A = \dfrac{{31}}{4} \Leftrightarrow \dfrac{2}{m} = - \dfrac{3}{2}\\
\Rightarrow m = \dfrac{{ - 4}}{3}\left( {tmdk} \right)
\end{array}$
