Đáp án: $(x,y)\in\{(2017, 2019), (2017, 2017)\}$
Giải thích các bước giải:
Ta có:
$|y-2018|+1\ge 0+1=1$
$\to \dfrac{2}{|y-2018|+1}\le\dfrac21=2$
$\to 3|x-2017|+1\le 2$
Mà $x\in Z\to 3|x-2017|+1\in Z$
Lại có $3|x-2017|+1\ge 3\cdot 0+1=1$
$\to 1\le 3|x-2017|+1\le 2$
Vì $3|x-2017|+1$ chia $3$ dư $1$
$\to 3|x-2017|+1=1$
$\to 3|x-2017|=0$
$\to |x-2017|=0$
$\to x-2017=0$
$\to x=2017$
$\to \dfrac{2}{|y-2018|+1}=3\cdot 0+1=1$
$\to |y-2018|+1=2$
$\to |y-2018|=1$
$\to y-2018=1\to y=2019$
Hoặc $y-2018=-1\to y=2017$
