Đáp án:
$\begin{array}{l}
a){x^2} + 2\left( {2m - 1} \right).x + 3\left( {{m^2} - 1} \right) = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow {\left( {2m - 1} \right)^2} - 3\left( {{m^2} - 1} \right) \ge 0\\
\Rightarrow 4{m^2} - 4m + 1 - 3{m^2} + 3 \ge 0\\
\Rightarrow {m^2} - 4m + 4 \ge 0\\
\Rightarrow {\left( {m - 2} \right)^2} \ge 0
\end{array}$
Vậy phương trình có nghiệm với mọi m
b) Pt có 2 nghiệm thì:
$\begin{array}{l}
m \ne 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\left( {2m - 1} \right)\\
{x_1}{x_2} = 3\left( {{m^2} - 1} \right)
\end{array} \right.\\
\Rightarrow 1 - 2m = \dfrac{{{x_1} + {x_2}}}{2}\\
\Rightarrow m = \dfrac{1}{2} - \dfrac{{{x_1} + {x_2}}}{4}\\
\Rightarrow {x_1}{x_2} = 3.{\left( {\dfrac{1}{2} - \dfrac{{{x_1} + {x_2}}}{4}} \right)^2} - 3\\
\Rightarrow {x_1}{x_2} = 3.\dfrac{{{{\left( {{x_1} + {x_2}} \right)}^2} - 4\left( {{x_1} + {x_2}} \right) + 4}}{{16}} - 3\\
Vậy\,\dfrac{3}{{16}}{\left( {{x_1} + {x_2}} \right)^2} - \dfrac{3}{4}\left( {{x_1} + {x_2}} \right) - \dfrac{9}{4} = {x_1}{x_2}
\end{array}$
