Đáp án:
$\begin{array}{l}
\frac{{{m^2} - 1}}{{{m^2} + 1}} = \frac{{{m^2} + 1 - 2}}{{{m^2} + 1}} = 1 - \frac{2}{{{m^2} + 1}}\\
do\,:{m^2} + 1 \ge 1\forall m\\
\Rightarrow \frac{2}{{{m^2} + 1}} \le 2\forall x\\
\Rightarrow - \frac{2}{{{m^2} + 1}} \ge - 2\forall x\\
\Rightarrow 1 - \frac{2}{{{m^2} + 1}} \ge - 1\forall x\\
\Rightarrow GTNN\,la\, - 1 \Leftrightarrow m = 0
\end{array}$