Đáp án:
$x=-2018$
Giải thích các bước giải:
$\dfrac{x-2}{505}+\dfrac{x+3}{2015}=\dfrac{x+6}{1006}+\dfrac{x+506}{504}$
$⇔ \dfrac{x-2}{505}+4+\dfrac{x+3}{2015}+1=\dfrac{x+6}{1006}+2+\dfrac{x+506}{504}+3$
$⇔ \dfrac{x+2018}{505}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{1006}+\dfrac{x+2018}{504}$
`⇔ (x+2018)(\frac{1}{505}+\frac{1}{2015}-\frac{1}{1006}-\frac{1}{504})=0`
$⇔ \left[ \begin{array}{l}x=-2018(tm)\\\dfrac{1}{505}+\dfrac{1}{2015}-\dfrac{1}{1006}-\dfrac{1}{504}=0(l)\end{array} \right.$
Vậy $x=-2018$
