\(\left(1+\frac{1}{11}\right)\cdot\left(1+\frac{1}{10}\right)\cdot\left(1+\frac{1}{9}\right)\cdot.........................................\left(1+\frac{1}{2}\right)\)
dấu . ở trên là nhân
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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:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
AT
12 tháng 2 2017
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
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
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
27 tháng 6 2019
\(\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{779}{780}\)
\(=\frac{4}{6}\cdot\frac{10}{12}\cdot\cdot\cdot\frac{1578}{1560}\)
\(=\frac{1\cdot4}{2\cdot3}\cdot\frac{2\cdot5}{3\cdot4}\cdot\cdot\cdot\frac{38\cdot41}{39\cdot40}\)
\(=\frac{\left(1\cdot4\right)\cdot\left(2\cdot5\right)\cdot\cdot\cdot\left(38\cdot41\right)}{\left(2\cdot3\right)\cdot\left(3\cdot4\right)\cdot\cdot\cdot\left(39\cdot40\right)}\)
\(=\frac{\left(1\cdot2\cdot\cdot\cdot38\right)\cdot\left(4\cdot5\cdot\cdot\cdot41\right)}{\left(2\cdot3\cdot\cdot\cdot39\right)\cdot\left(3\cdot4\cdot\cdot\cdot40\right)}\)
\(=\frac{1\cdot41}{39\cdot3}\)
\(=\frac{41}{117}\)
\(\left(1+\frac{1}{11}\right)\cdot\left(1+\frac{1}{10}\right)\cdot\left(1+\frac{1}{9}\right)\cdot.........................................\left(1+\frac{1}{2}\right)\)
\(=\frac{12}{11}.\frac{11}{10}.\frac{10}{9}....\frac{3}{2}\)
\(=\frac{12.11.10....3}{11.10.9....2}\)
\(=\frac{12}{2}=6\)
= \(\frac{12}{11}.\frac{11}{10}.....\frac{3}{2}=\frac{12}{2}=6\)