$M= \dfrac{x^4 -1 - (x^4 -x^2 +1)}{(x^2 +1)(x^4 -x^2+1)}. \dfrac{x^4 (x^2 +1) + (1-x^2)(1+x^2)}{x^2+1}$
$=\dfrac{x^2 -2}{(x^2+1)(x^4-x^2+1)}. (x^4 +1-x^2)= \dfrac{x^2-2}{x^2+1}$
Ta có $M=\dfrac{x^2 +1 - 3}{x^2+1} = 1-\dfrac{3}{x^2+1}$
Có $x^2+1 \ge 1 \Rightarrow \dfrac{3}{x^2+1} \le 3 \Rightarrow -\dfrac{3}{x^2+1} \ge- 3$
$\Rightarrow 1-\dfrac{3}{x^2+1} \ge 1-3=-2$
Vậy $\min M = -2$ khi $x=0$