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\(N=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+..+\frac{10}{1993.2003}\right)\)
\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)=\frac{3}{10}.\frac{2000}{6009}=\frac{200}{2003}\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{3}{13.23}\)\(+\)\(\frac{3}{23.33}\)\(+...+\)\(\frac{3}{1993.2003}\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}.\frac{1990}{26039}\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{597}{26039}\)
\(N=\)\(\frac{200}{2003}\)
\(=\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.99}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{99}\right)-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}-\frac{1}{2}\cdot\frac{49}{100}\)
\(=\frac{1}{2}\left(\frac{98}{99}-\frac{49}{100}\right)=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
Ta thấy các phân số trên khi quy đồng mẫu số chứa lũy thừa của 2 với số mũ cao nhất là 24
Như vậy, các phân số trên khi quy đồng mẫu số sẽ có tử chẵn, chỉ có phân số 1/16 có tử lẻ
=> tổng trên có tử lẻ, mẫu chẵn, không là số nguyên (đpcm)
C` cách 2 nhưng dài hơn
a) 75 - 3.(2/13+1/17-1/19) 3.[25-(2/13+1/17-1/19)]
-------------------------------------- = ---------------------------------------------- = 3/11
275-11.(2/13+1/17-1/19) 11.[25-(2/13+1/17-1/19)]
a) \(\frac{75-\frac{6}{13}+\frac{3}{17}-\frac{3}{19}}{275-\frac{22}{13}+\frac{11}{17}-\frac{11}{19}}=\frac{75-3.\left(\frac{2}{13}+\frac{1}{17}-\frac{1}{19}\right)}{275-11.\left(\frac{2}{13}+\frac{1}{17}-\frac{1}{19}\right)}\)
\(=\frac{75-3}{275-11}\)
\(=\frac{72}{264}=\frac{3}{11}\)
b) \(\frac{2}{3.5}+\frac{7}{5.12}+\frac{9}{4.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{12}+\frac{3}{52}\)
\(=\frac{1}{3}-\frac{1}{12}+\frac{3}{52}\)
\(=\frac{1}{4}+\frac{3}{52}=\frac{4}{13}\)
\(\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}+\frac{15}{31.46}+\frac{18}{46.64}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}+\frac{1}{31}-\frac{1}{46}+\frac{1}{46}-\frac{1}{64}\)
\(=1-\frac{1}{64}=\frac{63}{64}\)
Bạn ko hiểu chỗ nào là phải hỏi mình ngay nhé!
mấy bài này dễ lắm chỉ cần tính trong ngoặc trước, sau đó chỉ cần tính những hỗn số bên ngoài là xong
\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)
\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)
\(=\frac{1}{13}+\frac{597}{26039}\)
\(=\frac{200}{2003}\)
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