S=\(\frac{1}{10}\)+ \(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)

S=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)

S= \(\frac{1}{3}\).(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))

S= \(\frac{1}{3}\).(\(\frac{1}{2}\)-\(\frac{1}{20}\))

S= \(\frac{1}{3}\).\(\frac{9}{20}\)

S=\(\frac{3}{20}\)

chắc chắn nhé

\(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3S=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-....-\frac{1}{20}\)

\(3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)

\(\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)

a) \(\frac{1}{10}-\frac{1}{40}-\frac{1}{88}-\frac{1}{154}-\frac{1}{238}-\frac{1}{340}\)

\(=\frac{1}{10}-\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{20}\right)\)

\(=\frac{1}{10}-\frac{1}{3}.\frac{3}{20}\)

\(=\frac{1}{10}-\frac{1}{20}=\frac{2}{20}-\frac{1}{20}=\frac{1}{20}\)

\(A=\frac{3}{3}.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right)\)

\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right)\)

\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)

\(A=\frac{1}{3}.\frac{9}{20}\)

\(A=\frac{3}{20}\)

\(A=\frac{1}{2\times5}+\frac{1}{5\times8}+...+\frac{1}{17\times20}\)

\(A\times3=\frac{3}{2\times5}+\frac{3}{5\times8}+...+\frac{3}{17\times20}\)

\(A\times3=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(A\times3=\frac{1}{2}-\frac{1}{20}\)

\(A\times3=\frac{9}{20}\)

\(A=\frac{3}{20}\)

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(\Rightarrow A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(\Rightarrow3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow3A=\frac{1}{2}-\frac{1}{20}\)

\(\Rightarrow3A=\frac{9}{20}\)

\(\Rightarrow A=\frac{3}{20}\)

Bạn tách mẫu số ra kiểu 2 x 5 

                                    5 x 8

                                    ........

Cứ như thế

Sau đó rút gọn

Thực hiện một phép tính nữa

Vậy là ra kết quả

\(A=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

\(3A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)

\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\)

\(3A=\frac{1}{2}-\frac{1}{20}\)

\(3A=\frac{9}{20}\)

\(\Rightarrow A=\frac{3}{20}\)

\(\frac{1}{10}\)+\(\frac{1}{40}\)+\(\frac{1}{88}\)+\(\frac{1}{154}\)+\(\frac{1}{238}\)+\(\frac{1}{340}\)

=\(\frac{1}{2.5}\)+\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+\(\frac{1}{14.17}\)+\(\frac{1}{17.20}\)

=\(\frac{1}{3}\)(\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+\(\frac{3}{17.20}\))

=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+\(\frac{1}{17}\)-\(\frac{1}{20}\))

=\(\frac{1}{3}\)(\(\frac{1}{2}\)-\(\frac{1}{20}\))

=\(\frac{1}{3}\).\(\frac{9}{20}\)

=\(\frac{3}{20}\)

Ta có: S = 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340

=> S = 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 +1/14.17 +1/17.20
Nhân 2 vế với 3 và áp dụng công thức tách 1 phân số thành hiệu 2 phân số: x/n.(n + x) = 1/n - 1/(n + x)
=> 3.S = 3.(1/2.5 + 1/5.8 + 1/8.11 +1/11.14 +1/14.17 +1/17.20)
=> 3.S = 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 +3/14.17 +3/17.20
=> 3.S = 1/2 - 1/ 5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + 1/14 - 1/17 + 1/17 -1/20
=> 3.S = 1/2 - 1/20
=> 3.S = 9/20
=> S = 3/20

thanks nhiều nhak

Ta có:\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{340}=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{17.20}\)

=        \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{17}-\frac{1}{20}\right)=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)

\(=\frac{3}{20}=0,15\)

Đầu buồi mày thằng ngủ

Cái này biết nhưng quên mất công thức rùi

sorry nha!

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\)

= \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\)

= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right)\)

= \(\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)

= \(\frac{1}{3}.\frac{9}{20}\)

= \(\frac{3}{20}\)

Chú Ý dấu "." là dấu nhân nha

Ta có 

A=\(\frac{1}{10}\)+\(\frac{1}{40}\)+.....+\(\frac{1}{340}\)

=\(\frac{1}{2x5}\)+\(\frac{1}{5x8}\)+.....+\(\frac{1}{17x20}\)

=>3xA=\(\frac{3}{2x5}\)+\(\frac{3}{5x8}\)+.....+\(\frac{3}{17x20}\)

3xA=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+.....+\(\frac{1}{17}\)+\(\frac{1}{20}\)

3xA=\(\frac{1}{2}\)-\(\frac{1}{20}\)

3xA=\(\frac{9}{20}\)

=>A=\(\frac{9}{20}\):3

A=\(\frac{3}{20}\)

Vậy A=\(\frac{3}{20}\)

Ta có:

\(Coi\)  \(A=\left(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}+\frac{1}{340}\right).x=1\frac{1}{5}\)

               \(=\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+\frac{1}{14.17}+\frac{1}{17.20}\right).x=\frac{6}{5}\)

  \(\Rightarrow3A=\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\right).x=\frac{6}{5}.3=\frac{18}{5}\) \(=\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}\right).x\)

\(=\left(\frac{1}{2}-\frac{1}{20}\right).x\)

\(=\frac{9}{20}.x=\frac{18}{5}\)

\(\Rightarrow x=\frac{18}{5}:\frac{9}{20}=8\)

Vậy \(x=8\).