Đáp án:
\[{A_{\min }} = - \frac{1}{4} \Leftrightarrow m = - \frac{5}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} - \left( {m + 2} \right)x - m - 3 = 0\\
\Leftrightarrow {x^2} + x - \left( {m + 3} \right)x - \left( {m + 3} \right) = 0\\
\Leftrightarrow x\left( {x + 1} \right) - \left( {m + 3} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x - \left( {m + 3} \right)} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = m + 3
\end{array} \right.\\
A = - {x_1}^2{x_2} - {x_1}{x_2}^2 = - {x_1}{x_2}\left( {{x_1} + {x_2}} \right)\\
= \left( {m + 3} \right).\left( {m + 2} \right) = {m^2} + 5m + 6 = \left( {{m^2} + 2m.\frac{5}{2} + \frac{{25}}{4}} \right) - \frac{1}{4}\\
= {\left( {m + \frac{5}{2}} \right)^2} - \frac{1}{4} \ge - \frac{1}{4},\,\;\;\forall m
\end{array}\)
Vậy \({A_{\min }} = - \frac{1}{4} \Leftrightarrow m = - \frac{5}{2}\)
