Đáp án:
$\begin{array}{l}
{x^2} - 2\left( {m + 1} \right)x - m - 4 = 0\\
\Delta ' = {\left( {m + 1} \right)^2} + m + 4\\
= {m^2} + 2m + 1 + m + 4\\
= {m^2} + 3m + 5 > 0\forall x\\
\Rightarrow pt\,có\,2\,nghiệm\,pb\,\forall x\\
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 2\\
{x_1}{x_2} = - m - 4
\end{array} \right.\\
\Rightarrow M = \left| {{x_1} - {x_2}} \right|\\
\Rightarrow {M^2} = {\left( {{x_1} - {x_2}} \right)^2} = {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}\\
= {\left( {2m + 2} \right)^2} - 4.\left( { - m - 4} \right)\\
= 4{m^2} + 8m + 4 + 4m + 16\\
= 4{m^2} + 12m + 20\\
= 4\left( {{m^2} + 3m + 5} \right)\\
= 4\left( {{m^2} + 2.m.\frac{3}{2} + \frac{9}{4}} \right) + 11\\
= 4{\left( {m + \frac{3}{2}} \right)^2} + 11 \ge 11\forall x\\
\Rightarrow {M^2} \ge 11\\
\Rightarrow M \ge \sqrt {11} \\
\Rightarrow GTNN:M = \sqrt {11}
\end{array}$
