Tính \(P=\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot...\cdot\left(1-\frac{1}{1326}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NH
3 tháng 3 2020
\(\left[1-\frac{1}{21}\right]\times\left[1-\frac{1}{28}\right]\times\left[1-\frac{1}{36}\right]\times...\times\left[1-\frac{1}{1326}\right]\)
\(=\frac{20}{21}\times\frac{27}{28}\times\frac{35}{36}\times...\times\frac{1325}{1326}\)
\(=\frac{40}{42}\times\frac{54}{56}\times\frac{70}{72}\times...\times\frac{2650}{2652}\)
\(=\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times\frac{7\times10}{8\times9}\times...\times\frac{50\times53}{51\times52}\)
\(=\frac{5\times6\times7\times...\times50}{6\times7\times8\times...\times51}\times\frac{8\times9\times10\times...\times53}{7\times8\times9\times...\times52}\)
\(=\frac{5}{51}\times\frac{53}{7}\)
\(=\frac{265}{357}\)
T
26 tháng 7 2019
#)Giải :
\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{210}\right)=\frac{14}{15}\times\frac{20}{21}\times\frac{27}{28}\times...\times\frac{209}{210}\)
\(=\frac{28}{30}\times\frac{40}{42}\times\frac{54}{56}\times...\times\frac{418}{420}=\frac{4\times7}{5\times6}\times\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times...\times\frac{19\times22}{20\times21}\)
\(=\frac{4\times5\times6\times...\times19}{5\times6\times7\times...\times20}\times\frac{7\times8\times9\times...\times22}{6\times7\times8\times...\times21}=\frac{4}{20}\times\frac{22}{6}=\frac{11}{15}\)
26 tháng 7 2019
\(\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).....\left(1-\frac{1}{210}\right)\)
\(=\left(\frac{15}{15}-\frac{1}{15}\right).\left(\frac{21}{21}-\frac{1}{21}\right).\left(\frac{28}{28}-\frac{1}{28}\right).....\left(\frac{210}{210}-\frac{1}{210}\right)\)
\(=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}....\frac{209}{210}\)
\(=\frac{2.7}{3.5}.\frac{5.4}{7.3}.\frac{3.9}{4.7}....\frac{11.19}{21.10}\)
\(=\frac{2}{3}.\frac{19}{10}\)
\(=\frac{19}{15}\)
T
13 tháng 7 2019
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
26 tháng 6 2016
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
CM
24 tháng 7 2018
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
24 tháng 7 2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
