Đáp án:
$\begin{array}{l}
a)M = \left( {\frac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \frac{1}{{{x^2} + 1}}} \right)\left( {{x^4} + \frac{{1 - {x^4}}}{{1 + {x^2}}}} \right)\\
= \frac{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - \left( {{x^4} - {x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}.\frac{{{x^4}\left( {1 + {x^2}} \right) + 1 - {x^4}}}{{1 + {x^2}}}\\
= \frac{{{x^4} - 1 - {x^4} + {x^2} - 1}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}.\frac{{{x^4} + {x^6} + 1 - {x^4}}}{{1 + {x^2}}}\\
= \frac{{{x^2} - 2}}{{{x^6} + 1}}.\frac{{{x^6} + 1}}{{{x^2} + 1}}\\
= \frac{{{x^2} - 2}}{{{x^2} + 1}}\\
b)M = \frac{{{x^2} - 2}}{{{x^2} + 1}} = \frac{{{x^2} + 1 - 3}}{{{x^2} + 1}} = 1 - \frac{3}{{{x^2} + 1}}\\
Do:{x^2} + 1 \ge 1\forall x\\
\Rightarrow \frac{3}{{{x^2} + 1}} \le 3\forall x\\
\Rightarrow - \frac{3}{{{x^2} + 1}} \ge - 3\\
\Rightarrow 1 - \frac{3}{{{x^2} + 1}} \ge - 2\\
\Rightarrow M \ge - 2\\
\Rightarrow GTNN:M = - 2 \Leftrightarrow x = 0
\end{array}$
