b,
\(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}\)
\(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
=\(\frac{1}{4}-\frac{1}{10}=\frac{3}{20}\)
Bạn k rồi mình làm câu a cho