Giải thích các bước giải:
\(\begin{array}{l}
2{x^2} + (2m - 1)x + m - 1 = 0\\
a.m = 0\\
Pt \to 2{x^2} - x - 1 = 0\\
\to \left( {x - 1} \right)\left( {2x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \frac{1}{2}
\end{array} \right.
\end{array}\)
b. Để pt có 2 nghiệm phân biệt
⇔Δ>0
\(\begin{array}{l}
\to 4{m^2} - 4m + 1 - 8\left( {m - 1} \right) > 0\\
\to 4{m^2} - 12m + 9 > 0\\
\to {\left( {2m - 3} \right)^2} > 0\\
\Leftrightarrow m \ne \frac{3}{2}
\end{array}\)
\(\begin{array}{l}
\to \left\{ \begin{array}{l}
{x_1} = \frac{{ - 2m + 1 + \sqrt {{{\left( {2m - 3} \right)}^2}} }}{{2.2}}\\
{x_2} = \frac{{ - 2m + 1 - \sqrt {{{\left( {2m - 3} \right)}^2}} }}{{2.2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{x_1} = \frac{{ - 2m + 1 + 2m - 3}}{4} = - \frac{1}{2}\\
{x_2} = \frac{{ - 2m + 1 - 2m + 3}}{4} = \frac{{ - 4m + 4}}{4} = 1 - m
\end{array} \right.\\
Do:3{x_1} - 4{x_2} = 11\\
\to 3.\left( { - \frac{1}{2}} \right) - 4.\left( {1 - m} \right) = 11\\
\to \frac{{ - 3}}{2} - 4 + 4m = 11\\
\to m = \frac{{33}}{8}\left( {TM} \right)
\end{array}\)
