Giải thích các bước giải:
a.Ta có:
$2019|x-2019|+(x-2019)^2=2018|2019-x|$
$\to 2019|x-2019|+(x-2019)^2=2018|-(2019-x)|$
$\to 2019|x-2019|+(x-2019)^2=2018|-2019+x|$
$\to 2019|x-2019|+(x-2019)^2=2018|x-2019|$
$\to |x-2019|+(x-2019)^2=0$
Mà $|x-2019|\ge 0, (x-2019)^2\ge 0$
$\to |x-2019|+(x-2019)^2\ge 0$
Dấu = xảy ra khi $x-2019=0\to x=2019$
b.Ta có:
$2x+\dfrac17=\dfrac1y$
$\to 2xy+\dfrac{y}{7}=1$
$\to 14xy+y=7$
$\to y(14x+1)=7$
$\to (y,14x+1)$ là cặp ước của $7$
Mà $y\in Z^+$
$\to (y,14x+1)\in\{(7,1), (1,7)\}$
$\to (y,14x)\in\{(7,0), (1,6)\}$
$\to (y,x)\in\{(7,0), (1,\dfrac37)\}$
$\to (x,y)=(0,7)$
