Lời giải:
\(\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+...+\frac{1}{342}+\frac{1}{380}=\frac{1}{10\times 11}+\frac{1}{11\times 12}+...+\frac{1}{19\times 20}\)
\(=\frac{11-10}{10\times 11}+\frac{12-11}{11\times 12}+...+\frac{20-19}{19\times 20}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{10}-\frac{1}{20}=\frac{10}{200}=\frac{1}{20}\)
Đáp án A.