Đáp án:
$\begin{array}{l}
a){x^2} - \left( {2m - 1} \right).x - m + {m^2} = 0\\
\Rightarrow \Delta = {\left( {2m - 1} \right)^2} - 4\left( { - m + {m^2}} \right)\\
= 4{m^2} - 4m + 1 + 4m - 4{m^2}\\
= 1 > 0
\end{array}$
Vậy pt có 2 nghiệm phân biệt với mọi m
$\begin{array}{l}
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 1\\
{x_1}{x_2} = - m + {m^2}
\end{array} \right.\\
{x_1}\left( {1 - 2{x_2}} \right) + {x_2}\left( {1 - 2{x_1}} \right) = ??\\
\Rightarrow {x_1} + {x_2} - 4\left( {{x_1}{x_2}} \right) = ??\\
\Rightarrow 2m - 1 - 4\left( { - m + {m^2}} \right) = ??\\
\Rightarrow - 4{m^2} + 6m - 1 = ??\\
c)\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 1 \Rightarrow m = \dfrac{{{x_1} + {x_2} + 1}}{2}\\
{x_1}{x_2} = - m + {m^2}
\end{array} \right.\\
{x_1}{x_2} = - \dfrac{{{x_1} + {x_2} + 1}}{2} + {\left( {\dfrac{{{x_1} + {x_2} + 1}}{2}} \right)^2}\\
\Rightarrow {x_1}{x_2} = - \dfrac{{{x_1} + {x_2}}}{2} - \dfrac{1}{2} + {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2}\\
+ {x_1} + {x_2} + \dfrac{1}{4}\\
\Rightarrow {x_1}{x_2} - {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2} - \dfrac{{{x_1} + {x_2}}}{2} + \dfrac{1}{4} = 0
\end{array}$
