Đáp án:
$\begin{array}{l}
a){x^2} - 3x - m + 1 = 0\\
Khi:m = 2\\
\Leftrightarrow {x^2} - 3x - 2 + 1 = 0\\
\Leftrightarrow {x^2} - 3x - 1 = 0\\
\Leftrightarrow {x^2} - 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{9}{4} - 1 = 0\\
\Leftrightarrow {\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{13}}{4}\\
\Leftrightarrow x = \dfrac{{3 \pm \sqrt {13} }}{2}\\
Vậy\,x = \dfrac{{3 \pm \sqrt {13} }}{2}\,khi:m = 2\\
b)\Delta = 0\\
\Leftrightarrow {3^2} - 4\left( { - m + 1} \right) = 0\\
\Leftrightarrow 9 + 4m - 4 = 0\\
\Leftrightarrow m = - \dfrac{5}{4}\\
Vậy\,m = - \dfrac{5}{4}\\
c)\Delta > 0\\
\Leftrightarrow {3^2} - 4\left( { - m + 1} \right) > 0\\
\Leftrightarrow 4m + 5 > 0\\
\Leftrightarrow m > \dfrac{{ - 5}}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = - m + 1
\end{array} \right.\\
Khi:2{x_1} + {x_2} = 4\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
2{x_1} + {x_2} = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} = 1\\
{x_2} = 2
\end{array} \right.\\
\Leftrightarrow 1.2 = - m + 1\\
\Leftrightarrow m = - 1\left( {tmdk} \right)\\
Vậy\,m = - 1
\end{array}$
