Bài giải:

$\frac{x-15}{100}+\frac{x-10}{105}+\frac{x-5}{110}=\frac{x-100}{15}+\frac{x-105}{10}+\frac{x-110}{5}$

$\Leftrightarrow \left( \frac{x-15}{100}-1 \right)+\left( \frac{x-10}{105}-1 \right)+\left( \frac{x-5}{110}-1 \right)=\left( \frac{x-100}{15}-1 \right)+\left( \frac{x-105}{10}-1 \right)+\left( \frac{x-110}{5}-1 \right)$

$\Leftrightarrow \frac{x-115}{100}+\frac{x-115}{105}+\frac{x-115}{110}=\frac{x-115}{15}+\frac{x-115}{10}+\frac{x-115}{5}$

$\Leftrightarrow (x-115)\left( \frac{1}{100}+\frac{1}{105}+\frac{1}{110}-\frac{1}{15}-\frac{1}{10}-\frac{1}{5} \right)=0$

Mà: $\frac{1}{100}+\frac{1}{105}+\frac{1}{110}-\frac{1}{15}-\frac{1}{10}-\frac{1}{5}\ne 0$ nên x – 115 = 0

$\Leftrightarrow $ x = 115

Vậy nghiệm của phương trình là: x = 115