$\frac{x-2004}{15}+$ $\frac{x-1995}{12}+$ $\frac{x-1989}{10}+$ $\frac{x-1987}{8}=10$
$⇔(\frac{x-2004}{15}-1)+($ $\frac{x-1995}{12}-2)+($ $\frac{x-1989}{10}-3)+($ $\frac{x-1987}{8}=0$
$⇔\frac{x-2019}{15}+$ $\frac{x-2019}{12}+$ $\frac{x-2019}{10}+$ $\frac{x-2019}{8}=0$
$⇔(x-2019).($ $\frac{1}{15}+$ $\frac{1}{12}+$ $\frac{1}{10}+$ $\frac{1}{8})=0$ (vì: $\frac{1}{15}+$ $\frac{1}{12}+$ $\frac{1}{10}+$ $\frac{1}{8}\neq0$)
$⇔x-2019=0$
$⇔x=2019$
Vậy $S=\{2019\}$
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