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Đặt \(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(\Leftrightarrow A=\frac{2}{5}.\frac{505}{2036}\)

\(\Leftrightarrow A=\frac{101}{1018}\)

~ Hok tốt ~

#)Giải :

\(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(A=\frac{2}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{504.509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\times\frac{505}{2036}\)

\(A=\frac{101}{1018}\)

\(A=\frac{1}{7}\left[\frac{1}{2}-\frac{1}{9}+...+\frac{1}{252}-\frac{1}{509}\right]\)

\(A=\frac{1}{7}.\left[\frac{1}{2}-\frac{1}{509}\right]\)

\(A=\frac{1}{7}.\left[\frac{507}{1018}\right]=\frac{507}{7126}\)

mk nghĩ là vậy đó, ủng hộ mk nha

Ta có:

\(A=\frac{1}{2.9}+\frac{1}{9.7}+\frac{1}{7.19}+...+\frac{1}{252.509}\)

\(A=\frac{2}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{1}{504.509}\right)\)

\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}.\left(\frac{1}{4}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}.\frac{505}{2036}\)

\(A=\frac{101}{1018}.\)

Vậy \(A=\frac{101}{1018}.\)

Chúc bạn học tốt!

trời

anh ơi anh anh dẹp cho em nhờ

bạn viết sai đề rồi . phải là 7 . 9 chứ

\(A=\frac{2}{4.9}+\frac{2}{9.14}+\frac{2}{14.19}+...+\frac{2}{504.509}\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{504}-\frac{1}{509}\right)\)

\(A=\frac{2}{5}\left(\frac{1}{4}-\frac{1}{509}\right)=...\)

\(B=\frac{1.4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+...+\frac{100.103+1}{100.103}\)

\(B=1+\frac{1}{1.4}+1+\frac{1}{4.7}+...+1+\frac{1}{100.103}\)

\(B=34+\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(B=34+\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=34+\frac{1}{3}\left(1-\frac{1}{103}\right)=...\)

a)Ta có: 

A= 1/2.9 + 1/9.7 +...+1/252.509

  = 2/5.(5/4.9 + 5/9.14 + 5/14.19 +...+ 1/504.509)

  = 2/5.(1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/504 - 1/509)

  = 2/5.(1/4 - 1/509)

  = 101/1018

Vậy A = 101/1018

b)Ta có:

B= 1/10.9 +1/18.13 + 1/26.17 +...+ 1/802.405)

  = 1/4.(8/10.18 + 8/18.26 + 8/26.34 +...+ 8/802.810)

  = 1/4.(1/10 - 1/18 + 1/18 - 1/26 + 1/26 - 1/34 +...+ 1/802 - 1/810)

  = 1/4.(1/10 - 1/810)

  = 2/81

Vậy B= 2/81

Tk mình nha!!!

a,A=101/1018 b,B=2/81

\(\frac{n}{n}\)