a) \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(A=2\cdot\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{15\cdot16}\right)\)

\(A=2\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(A=2\cdot\left(\frac{1}{4}-\frac{1}{16}\right)=2\cdot\frac{3}{16}=\frac{3}{8}\)

b) \(B=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)

\(B=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+...+\frac{5}{700}\)

\(B=\frac{5}{3}\cdot\left(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{25\cdot28}\right)\)

\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)

\(B=\frac{5}{3}\cdot\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}\cdot\frac{3}{14}=\frac{5}{14}\)

Ta có: \(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(\Rightarrow B=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\)

\(\Rightarrow B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)\)

\(\Rightarrow B=2.\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)

Vậy \(B=\frac{3}{8}\)

nha m.n

                                    \(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+.....+\frac{1}{120}\)

                                     \(B=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{240}\right)\)

                                    \(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{15.16}\right)\)

                                    \(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+......+\frac{1}{15}-\frac{1}{16}\right)\)

                                    \(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

                                      \(B=2.\frac{3}{16}\)

                                    \(B=\frac{3}{8}\)

                                   Vậy \(B=\frac{3}{8}\)

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)

\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(A=2.\frac{3}{16}=\frac{3}{8}\)

\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)

\(=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(=2.\frac{3}{16}=\frac{3}{8}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(A=1-\frac{1}{6}\)

\(A=\frac{6}{6}-\frac{1}{6}\)

\(A=\frac{5}{6}\)

\(B=\frac{1}{10}+\frac{1}{15}+...+\frac{1}{120}\)

\(B=2.\left(\frac{1}{20}+\frac{1}{30}+...+\frac{1}{240}\right)\)

\(B=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+..+\frac{1}{15.16}\right)\)

\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(B=2.\left(\frac{4}{16}-\frac{1}{16}\right)\)

\(B=2.\frac{3}{16}\)

\(B=\frac{3}{8}\)

Chúc bạn học tốt !!! 

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{5.6}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{5}-\frac{1}{6}\)

\(A=\frac{1}{1}-\frac{1}{6}\)

\(A=\frac{5}{6}\)

\(B=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)

\(B=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)

\(B=\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{15.16}\)

\(B=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)

\(B=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)

\(B=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)

\(B=2.\frac{3}{16}\)

\(B=\frac{6}{16}=\frac{3}{13}\)