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 \(B=\frac{3}{8}\)

Có: \(N=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\)

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

\(=>N=\frac{2}{4\cdot5}+\frac{2}{5\cdot6}+\frac{2}{6\cdot7}+...+\frac{2}{15\cdot16}\)

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

\(=>N=\frac{2}{4}-\frac{2}{16}\)

\(=>N=\frac{1}{2}-\frac{1}{8}\)

\(=>N=\frac{8-2}{16}=\frac{6}{16}=\frac{3}{8}\)

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

Ta có : 

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

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

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

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

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

\(N=\frac{1}{2}-\frac{1}{8}\)

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

Vậy \(N=\frac{3}{8}\)

Chúc bạn học tốt ~ 

\(=\frac{2}{20}+\frac{2}{30}+...+\frac{2}{420}\)

\(=\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{20.21}\)

\(=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(=2\left(\frac{1}{4}-\frac{1}{21}\right)\)

\(=2\times\frac{17}{84}\)

\(=\frac{17}{72}\)

Ta có:

\(\frac{x}{2013}\)-\(\frac{1}{10}\)-\(\frac{1}{15}\)-\(\frac{1}{21}\)-...-\(\frac{1}{120}\)=\(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- (\(\frac{2}{20}\)+\(\frac{2}{30}\)+\(\frac{2}{42}\)+...+\(\frac{2}{240}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4.5}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{15.16}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.(\(\frac{1}{4}\)-\(\frac{1}{10}\)) = \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)- 2.\(\frac{3}{10}\)= \(\frac{5}{8}\)

=>\(\frac{x}{2013}\)= \(\frac{5}{8}\)+\(\frac{6}{10}\)= 1

=> \(x=2013\)

Vậy \(x=2013\)

2013 nha

khong hieu

Ta có:

\(\Rightarrow\frac{x}{2008}=1\)

\(\Rightarrow x=1.2008\)

\(\Rightarrow x=2008\)

Vậy \(x=2008.\)

Chúc bạn học tốt!