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PT
1
14 tháng 7 2016
\(A=\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).\left(1-\frac{1}{36}\right)...\left(1-\frac{1}{1326}\right)\)
\(A=\frac{20}{21}.\frac{27}{28}.\frac{35}{36}...\frac{1325}{1326}\)
\(A=\frac{40}{42}.\frac{54}{56}.\frac{70}{72}...\frac{2650}{2652}\)
\(A=\frac{5.8}{6.7}.\frac{6.9}{7.8}.\frac{7.10}{8.9}...\frac{50.53}{51.52}\)
\(A=\frac{5.6.7...50}{6.7.8...51}.\frac{8.9.10...53}{7.8.9...52}\)
\(A=\frac{5}{51}.\frac{53}{7}\)
\(A=\frac{265}{357}\)
NL
1
8 tháng 1 2017
\(A=xemlai\) chưa hưa hiểu Quy luật
\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)
\(B=\frac{2.100}{1.101}=\frac{200}{101}\)
DP
3 tháng 8 2017
\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot...\cdot\left(1-\frac{1}{1326}\right)\)
\(=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot...\cdot\frac{1325}{1326}\)
\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot...\cdot\frac{2650}{2652}\)
\(=\frac{5.8}{6.7}\cdot\frac{6.9}{7.8}\cdot\frac{7.10}{8.9}\cdot....\cdot\frac{50.53}{51.52}\)
\(=\frac{5\cdot6\cdot7\cdot.....\cdot50}{6\cdot7\cdot8\cdot....\cdot51}\cdot\frac{8\cdot9\cdot10\cdot....\cdot53}{7\cdot8\cdot9\cdot...\cdot52}=\frac{5}{51}\cdot\frac{53}{7}=\frac{265}{357}\)
15 tháng 7 2018
\(\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)\left(1-\frac{1}{36}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{1326}\right)\)
\(=\frac{20}{21}\cdot\frac{27}{28}\cdot\frac{35}{36}\cdot\cdot\cdot\cdot\cdot\frac{1325}{1326}\)
\(=\frac{40}{42}\cdot\frac{54}{56}\cdot\frac{70}{72}\cdot\cdot\cdot\cdot\cdot\cdot\frac{2650}{2652}\)
\(=\frac{5\cdot8}{6\cdot7}\cdot\frac{6\cdot9}{7\cdot8}\cdot\frac{7\cdot10}{8\cdot9}\cdot\cdot\cdot\cdot\cdot\frac{50\cdot53}{51\cdot52}\)
\(=\frac{\left(5\cdot8\right)\left(6\cdot9\right)\left(7\cdot10\right)\cdot\cdot\cdot\cdot\cdot\left(50\cdot53\right)}{\left(6\cdot7\right)\left(7\cdot8\right)\left(8\cdot9\right).....\left(51\cdot52\right)}\)
\(=\frac{\left(5\cdot6\cdot7\cdot\cdot\cdot\cdot\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot\cdot\cdot\cdot\cdot53\right)}{\left(6\cdot7\cdot8\cdot\cdot\cdot\cdot\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot\cdot\cdot\cdot\cdot52\right)}\)
\(=\frac{5\cdot53}{51\cdot7}=\frac{265}{357}\)