Cho phương trình (x-2013)/16+(x-1997)/11+(x-1992)/9+(x-1991)/7=10

* Hướng dẫn giải

$\begin{array}{l}\frac{x-2003}{16}+\frac{x-1997}{11}+\frac{x-1992}{9}+\frac{x-1991}{7}=10\\ \iff \left(\frac{x-2003}{16}-1\right)+\left(\frac{x-1997}{11}-2\right)+\left(\frac{x-1992}{9}-3\right)+\left(\frac{x-1991}{7}-4\right)=0\end{array}$

$\begin{array}{l}\iff \frac{x-2019}{16}+\frac{x-2019}{11}+\frac{x-2019}{9}+\frac{x-2019}{7}=0\\ \iff \left(x-2019\right)\left(\frac{1}{16}+\frac{1}{11}+\frac{1}{9}+\frac{1}{7}\right)=0\\ Do\frac{1}{16}+\frac{1}{11}+\frac{1}{9}+\frac{1}{7}>0\Rightarrow x-2019=0\iff x=2019S=\left\{2019\right\}\end{array}$