\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)

\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)

\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)

\(=\frac{1}{10}.\frac{2000}{2003}\)

\(=\frac{200}{2003}\)

Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)

\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)

\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)

\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)

\(\Rightarrow A=\frac{200}{2003}\)

\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)

\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)

\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)

\(=\frac{1}{13}+\frac{597}{26039}\)

\(=\frac{200}{2003}\)

Đặt A= 1/13 + 3/13.23 + 3/ 23.33 + ... + 3/1993.2003 

A- 1/13 = 3/13.23 + 3/ 23.33 + ... + 3/1993.2003 

10/3 ( A-1/3) =  10/3. (3/13.23 + 3/ 23.33 + ... + 3/1993.2003) 

10/3A - 10/9 = 10/13.23 + 10/ 23.33 + ... + 10/1993.2003 

10/3A - 10/9  = 1/13 - 1/23 + 1/23 - 1/33 +...+ 1/1993- 1/2003

10/3A = 1/13 - 1/2003 + 10/9

10/3 A= ? 

đến đây bn tự làm nha

10/3A - 10/9 = 1/13 

Ta loại \(\frac{1}{13}\)ra khỏi biểu thức để dễ tính hơn, sau đó cộng vào.

Gọi A là biểu thức \(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

Gấp A lên \(\frac{10}{3}\)lần, ta có :

\(A\times\frac{10}{3}=\frac{10}{3}\times\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)

\(A\times\frac{10}{3}=\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\)

\(A\times\frac{10}{3}=\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\)

\(A\times\frac{10}{3}=\frac{1}{13}-\frac{1}{2003}\)

\(A\times\frac{10}{3}=\frac{1990}{26039}\)

\(A=\frac{1990}{26039}\div\frac{10}{3}\)

\(A=\frac{597}{26039}\)

Biểu thức trên = \(\frac{597}{26039}+\frac{1}{13}=\frac{200}{2003}\)

 \(N=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)

\(=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+..+\frac{10}{1993.2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)

\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)=\frac{3}{10}.\frac{2000}{6009}=\frac{200}{2003}\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{3}{13.23}\)\(+\)\(\frac{3}{23.33}\)\(+...+\)\(\frac{3}{1993.2003}\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{2003}\right)\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}.\frac{1990}{26039}\right]\)

\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{597}{26039}\)

\(N=\)\(\frac{200}{2003}\)

dễ vãi lồn làm làm đéo j phá đê

\(B=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)

\(B=10.\left(\frac{1}{3.13}+\frac{1}{13.23}+....+\frac{1}{53.63}\right)\)

\(B=10.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}+\frac{1}{63}\right)\)

\(B=10.\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(B=10.\frac{20}{63}\)

\(B=\frac{200}{63}\)

Cảm ơn bn nha

\(\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}+\frac{7}{33.43}\)

\(=\frac{7}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+\frac{10}{33.43}\right)\)

\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+\frac{1}{33}-\frac{1}{43}\right)\)

\(=\frac{7}{10}\left(\frac{1}{3}-\frac{1}{43}\right)\)

\(=\frac{7}{10}\left(\frac{43}{129}-\frac{3}{129}\right)\)

\(=\frac{7}{10}.\frac{40}{129}\)

\(=\frac{28}{129}\)

mk làm đúng rồi nha, ko tin bấm thử máy tính

 7/3.13 + 7/13.23 + 7/23.33 + 7/33.43

= 7/10.(1/3-1/13+1/13-1/23+1/23-1/33+1/33-1/43)

= 7/10.(1/3-1/43)

= 7/10 . 14/43

= 49/215

Đặt A ta có :

\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times56}\)

\(A=\frac{7}{10}\times\left(\frac{10}{3\times13}+\frac{10}{13\times23}+...+\frac{10}{53\times56}\right)\)

\(A=\frac{7}{10}\times\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+...+\frac{1}{53}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}\times\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(A=\frac{7}{10}\times\frac{20}{63}\)

\(\Rightarrow A=\frac{2}{9}\)

phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))

\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)

b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)

(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)

=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)

=1/101+1/102+...+1/200

https://olm.vn/hoi-dap/question/60726.html

\(A=\frac{7}{3.13}+\frac{7}{13.23}+...+\frac{7}{53.63}\)

\(A=7.\left(\frac{1}{3.13}+\frac{1}{13.23}+...+\frac{1}{53.63}\right)\)

\(10A=7.\left(\frac{10}{3.13}+\frac{10}{13.23}+...+\frac{10}{53.63}\right)\)

\(10A=7.\left(\frac{1}{3}-\frac{1}{63}\right)\)

\(10A=7.\frac{20}{63}\)

\(10A=\frac{20}{9}\)

\(A=\frac{20}{9}:10\)

\(A=\frac{20}{9}.\frac{1}{10}\)

\(A=\frac{2}{9}\)

Vậy A=\(\frac{2}{9}\)

Chúc bạn học tốt~

A = \(\frac{7}{3.13}\)\(+\)\(\frac{7}{13.23}\)\(+\)\(\frac{7}{23.33}\)\(+\)..........  \(+\)\(\frac{7}{53.63}\)

\(\Rightarrow\)A = 7 . (\(\frac{1}{3.13}\)\(+\)\(\frac{1}{13.23}\)\(+\)\(\frac{1}{23.33}\)\(+\)...... \(\frac{1}{53.63}\))

\(\Rightarrow\)A = \(\frac{1}{10}\). 7 . ( \(\frac{1}{3}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{23}\)\(+\)\(\frac{1}{23}\)\(-\)\(\frac{1}{33}\)\(+\)....... \(+\)\(\frac{1}{53}\)\(-\)\(\frac{1}{63}\))

\(\Rightarrow\)A = \(\frac{7}{10}\). ( \(\frac{1}{3}\)\(-\)\(\frac{1}{63}\))

\(\Rightarrow\)A = \(\frac{7}{10}\). \(\frac{20}{63}\)

\(\Rightarrow\)A = \(\frac{2}{9}\)

Chúc các anh em học tốt !!!

A=1/3-1/13+1/13-1/23+1/23-1/33+.........+1/53-1/63

=1/3-1/63

=21/63-1/63

=20/63

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(2S=\frac{2.5}{3.13}+\frac{2.5}{13.23}+....+\frac{2.5}{83.93}\)

\(2S=\frac{10}{3.13}+\frac{10}{13.23}+.....+\frac{10}{83.93}\)

\(2S=\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\)

\(2S=\frac{1}{3}-\frac{1}{93}=\frac{30}{93}\)

\(S=\frac{30}{93}.\frac{1}{2}=\frac{15}{93}\)

Sửa đề:

\(S=\frac{5}{3.13}+\frac{5}{13.23}+.....+\frac{5}{83.93}\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{83}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\left(\frac{31}{93}-\frac{1}{93}\right)\)

\(S=\frac{1}{2}.\frac{10}{31}\)

\(S=\frac{5}{31}\)

bn lớp 5 ak