Tính : \(\left(1-\frac{1}{21}\right)\)\(\left(1-\frac{1}{28}\right)\)\(\left(1-\frac{1}{36}\right)\)..........\(\left(1-\frac{1}{1326}\right)\)
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12 tháng 2 2019
\(A=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right).....\left(1-\frac{1}{1326}\right)\)
\(A=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}......\frac{1325}{1326}\)
\(A=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}.....\frac{50.53}{51.52}\)
\(A=\frac{5.\left(6.7.....50\right)}{\left(6.7.....50\right).51}.\frac{53.\left(8.9.10.....52\right)}{7.\left(8.9.10.......52\right)}\)
\(A=\frac{5}{51}.\frac{53}{7}=\frac{265}{357}\)
29 tháng 6 2015
\(A=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}....\frac{1325}{1326}=\frac{40}{42}.\frac{54}{56}.\frac{70}{72}....\frac{2650}{2652}\)
\(A=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}....\frac{50.53}{51.52}=\frac{\left(5.6.7...50\right).\left(8.9.10...53\right)}{\left(6.7.8...51\right).\left(7.8.9...52\right)}=\frac{5.53}{51.7}=..\)
O
24 tháng 3 2018
\(A=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot...\cdot\frac{1325}{1326}\)
\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot...\cdot\frac{2650}{2652}\)
\(=\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot\frac{7\cdot10}{8\cdot9}\cdot...\cdot\frac{50\cdot53}{51\cdot52}\)
\(=\frac{5\cdot53}{7\cdot51}=\frac{265}{357}\)
6 tháng 7 2016
\(\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
\(=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}...\frac{1325}{1326}\)
\(=\frac{40}{42}.\frac{54}{56}.\frac{70}{72}...\frac{2650}{2652}\)
\(=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}...\frac{50.53}{51.52}\)
\(=\frac{5.6.7...50}{7.8.9...52}.\frac{8.9.10...53}{6.7.8...51}\)
\(=\frac{5.6}{51.52}.\frac{52.53}{6.7}\)
\(=\frac{5.52}{51.7}=\frac{260}{357}\)
Ủng hộ mk nha ^_-
Bạn alibaba nguyễn giải đúng rồi nhưng mình nghĩ cách này sẽ nhanh hơn :
Giải :
Đặt : \(A=\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)........\left(1-\frac{1}{1326}\right)\)
\(\Rightarrow A=\left(1-\frac{2}{6.7}\right)\left(1-\frac{2}{7.8}\right)\left(1-\frac{2}{8.9}\right).......\left(1-\frac{2}{51.52}\right)\)
\(\Rightarrow A=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}.........\frac{50.53}{51.52}\)
\(\Rightarrow A=\frac{\left(5.6.7......50\right)\left(8.9.10......53\right)}{\left(6.7.8.....51\right)\left(7.8.9......52\right)}\)
\(\Rightarrow A=\frac{5}{51}.\frac{53}{7}\)
\(\Rightarrow A=\frac{265}{357}\)
Vậy : \(\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)......\left(1-\frac{1}{1326}\right)=\frac{265}{357}\)
tớ nghĩ là bằng \(\frac{1}{1326}\)