\(\left(\frac{1}{38}-1\right)\cdot\left(\frac{1}{37}-1\right)\cdot\left(\frac{1}{36}-1\right)\cdot...\cdot\left(\frac{1}{2}-1\right)\)
mik dang can gap
mik se tick cho
THANKS
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
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:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
\(\left[1-\frac{1}{21}\right]\times\left[1-\frac{1}{28}\right]\times\left[1-\frac{1}{36}\right]\times...\times\left[1-\frac{1}{1326}\right]\)
\(=\frac{20}{21}\times\frac{27}{28}\times\frac{35}{36}\times...\times\frac{1325}{1326}\)
\(=\frac{40}{42}\times\frac{54}{56}\times\frac{70}{72}\times...\times\frac{2650}{2652}\)
\(=\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times\frac{7\times10}{8\times9}\times...\times\frac{50\times53}{51\times52}\)
\(=\frac{5\times6\times7\times...\times50}{6\times7\times8\times...\times51}\times\frac{8\times9\times10\times...\times53}{7\times8\times9\times...\times52}\)
\(=\frac{5}{51}\times\frac{53}{7}\)
\(=\frac{265}{357}\)
= (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
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
\(\left(\frac{1}{38}-1\right).\left(\frac{1}{37}-1\right).\left(\frac{1}{36}-1\right)....\left(\frac{1}{2}-1\right)\)
\(=-\left(1-\frac{1}{38}\right).\left(1-\frac{1}{37}\right)....\left(1-\frac{1}{2}\right)\)
\(=-\frac{37}{38}.\frac{36}{37}...\frac{1}{2}\)
\(=\frac{-1}{38}\)
\(\left(\frac{1}{38}-1\right)\left(\frac{1}{37}-1\right)\left(\frac{1}{36}-1\right).......\left(\frac{1}{2}-1\right)\)
\(=\frac{-37}{38}.\frac{-36}{37}.\frac{-35}{36}.......\frac{-2}{3}.\frac{-1}{2}\)
\(=\frac{-1}{38}\)