$\dfrac{x-15}{1863}+\dfrac{x-12}{1866}=\dfrac{x}{626}-1$
$↔️(\dfrac{x-15}{1863}-1)+(\dfrac{x-12}{1866}-1)-(\dfrac{x}{626}-3)=0$
$↔️\dfrac{x-15-1863}{1863}+\dfrac{x-12-1866}{1866}-\dfrac{x-626.3}{626}=0$
$↔️\dfrac{x-1878}{1863}+\dfrac{x-1878}{1866}-\dfrac{x-1878}{626}=0$
$↔️(x-1878)(\dfrac{1}{1863}+\dfrac{1}{1866}-\dfrac{1}{626})=0$
Do $\dfrac{1}{1863}+\dfrac{1}{1866}-\dfrac{1}{626} \ne 0$
$↔️x-1878=0$
$↔️x=1878$
Vậy $S=\{1878\}$
