Đáp án:

$\\$

`M = |x - 2002| + |x -2001|`

`↔ M = |x -2002| + |2001 - x|`

Áp dụng BĐT $|a| + |b| \geqslant |a + b|$

$↔ |x - 2002| + |2001 - x| \geqslant |x - 2002 + 2001 - x| = |1| = 1$

$↔ M \geqslant 1$

`↔ min M = 1`

Dấu "`=`" xảy ra khi :

$↔ (x - 2002) (2001 - x) \geqslant 0$

$↔$ \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x-2002 \geqslant 0\\2001 - x \geqslant 0\end{array} \right.\\ \left\{ \begin{array}{l}x-2002\leqslant 0 \\2001 - x \leqslant 0\end{array} \right.\end{array} \right.\)

`↔` \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x \geqslant 2002\\ x \leqslant 2001\end{array} \right . \text{(Vô lí)}\\ \left\{ \begin{array}{l}x\leqslant 2002 \\ x \geqslant2 001\end{array} \right. \text{(Luôn đúng)}\end{array}  \right.\)

`↔` $2001 \leqslant x \leqslant 2002$

Vậy `min M = 1 ↔` $2001 \leqslant x \leqslant 2002$