Đáp án:
$y=100; z=2021; \to -2020\le x \le 2022 $
Giải thích các bước giải:
Ta có:
$|x+2020|+|y-100|+|x-2022|+|z-2021|=4042$
$\to (|x+2020|+|x-2022|)+|y-100|+|z-2021|=4042$
Mà $(|x+2020|+|x-2022|)+|y-100|+|z-2021|=(|x+2020|+|2022-x|)+|y-100|+|z-2021|\ge |x+2020+2022-x|+0+0=4042$
$\to$ Dấu = xảy ra khi $(x+2020)(2022-x)\ge 0;y-100=z-2021=0$
+ $ y-100=z-2021=0$
$ <=> y=100, z=2021$
+ $(x+2020)(2022-x)\ge 0$
$ \to -2020 \le x \le 2022$
