Đáp án:
$\begin{array}{l}
4{x^2} + 2\left( {2m + 3} \right)x + {m^2} - 3m + 2 = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow {\left( {2m + 3} \right)^2} - 4.\left( {{m^2} - 3m + 2} \right) \ge 0\\
\Rightarrow 4{m^2} + 12m + 9 - 4{m^2} + 12m - 8 \ge 0\\
\Rightarrow 24m + 1 \ge 0\\
\Rightarrow m \ge - \dfrac{1}{{24}}\\
Theo\,Viet:{x_1}.{x_2} = \dfrac{{{m^2} - 3m + 2}}{4}\\
= \dfrac{{{{\left( {m - \dfrac{3}{2}} \right)}^2} - \dfrac{9}{4} + 2}}{2}\\
= \dfrac{{{{\left( {m - \dfrac{3}{2}} \right)}^2} - \dfrac{1}{4}}}{2}\\
Do:{\left( {m - \dfrac{3}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\Rightarrow \dfrac{{{{\left( {m - \dfrac{3}{2}} \right)}^2} - \dfrac{1}{4}}}{2} \ge \dfrac{{ - 1}}{8}\\
\Rightarrow {x_1}{x_2} \ge - \dfrac{1}{8}\\
\Rightarrow GTNN:{x_1}{x_2} = - \dfrac{1}{8} \Leftrightarrow m = \dfrac{1}{2}\left( {tmdk} \right)
\end{array}$
Vậy tích 2 nghiệm đạt GTNN bằng -1/8 khi m=1/2.
