Đáp án:
$\begin{array}{l}
38)\\
{x^2} - \left( {2m + 3} \right).x + m - 3 = 0\\
\Delta = {\left( {2m + 3} \right)^2} - 4m + 12\\
= 4{m^2} + 12m + 9 - 4m + 12\\
= 4{m^2} + 8m + 21\\
= 4\left( {{m^2} + 2m + 1} \right) + 17\\
= 4{\left( {m + 1} \right)^2} + 17 > 0\\
TheoViet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 3\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
A = \left| {{x_1} - {x_2}} \right|\\
= \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2}} \\
= \sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
= \sqrt {{{\left( {2m + 3} \right)}^2} - 4.m + 12} \\
= \sqrt {4{{\left( {m + 1} \right)}^2} + 17} \ge \sqrt {17} \\
\Rightarrow GTNN:A = \sqrt {17} \\
Khi:m = - 1\\
39)a)\\
a)Do:{x^2} - \left( {m - 1} \right).x - {m^2} + m - 2 = 0\\
a = 1;b = - \left( {m - 1} \right);c = - {m^2} + m - 2\\
\Rightarrow a.c = 1.\left( { - {m^2} + m - 2} \right)\\
= - \left( {{m^2} - m + 2} \right)\\
= - \left( {{m^2} - 2.m.\frac{1}{2} + \frac{1}{4} + \frac{7}{4}} \right)\\
= - {\left( {m - \frac{1}{2}} \right)^2} - \frac{7}{4} \le - \frac{7}{4} < 0\\
\Rightarrow a.c < 0
\end{array}$
Vậy pt luôn có 2 nghiệm trái dấu với mọi m
