Bài giải:

$\left( \frac{1}{1.51}+\frac{1}{2.52}+\frac{1}{3.53}+...+\frac{1}{10.60} \right)x=\left( \frac{1}{1.11}+\frac{1}{2.12}+\frac{1}{3.13}+...+\frac{1}{50.60} \right)$

$\Leftrightarrow \frac{1}{50}.\left( \frac{50}{1.51}+\frac{50}{2.52}+\frac{50}{3.53}+...+\frac{50}{10.60} \right)x=\frac{1}{10}\left( \frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{50.60} \right)$

$\Leftrightarrow \frac{1}{50}x.\left( \frac{1}{1}-\frac{1}{51}+\frac{1}{2}-\frac{1}{52}+\frac{1}{3}-\frac{1}{53}+...+\frac{1}{10}-\frac{1}{60} \right)=\frac{1}{10}\left( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{50}-\frac{1}{60} \right)$

$\Leftrightarrow \frac{x}{50}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60} \right) \right]=\frac{1}{10}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50} \right)-\left( \frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{60} \right) \right]$$\Leftrightarrow \frac{x}{50}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60} \right) \right]=\frac{1}{10}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)+\left( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{50} \right)-\left( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{50} \right)-\left( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60} \right) \right]$

$\Leftrightarrow \frac{x}{50}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60} \right) \right]=\frac{1}{10}\left[ \left( 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10} \right)-\left( \frac{1}{51}+\frac{1}{52}+...+\frac{1}{60} \right) \right]$ $\Leftrightarrow \frac{x}{50}=\frac{1}{10}$

$\Leftrightarrow $ x= 5

Tập nghiệm của phương trình là: S = {5}

Vậy đáp án đúng là: C