\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2017}\right)\left(1-\dfrac{1}{2018}\right)\)
\(=\dfrac{2-1}{2}.\dfrac{3-1}{3}.\dfrac{4-1}{4}...\dfrac{2017-1}{2017}.\dfrac{2018-1}{2018}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2016}{2017}.\dfrac{2017}{2018}\)
\(=\dfrac{1.2.3...2016.2017}{2.3.4...2017.2018}\)
\(=\dfrac{1}{2018}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2017.2018}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(=1-\dfrac{1}{2018}\)
\(=\dfrac{2018-1}{2018}\)
\(=\dfrac{2017}{2018}\)
\(D=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)
\(=\dfrac{3}{3.1.4}+\dfrac{3}{3.4.7}+\dfrac{3}{3.7.10}+...+\dfrac{3}{3.97.100}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}\left[\left(1-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{10}\right)+...+\left(\dfrac{1}{97}-\dfrac{1}{100}\right)\right]\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{100-1}{100}\)
\(=\dfrac{1}{3}.\dfrac{99}{100}\)
\(=\dfrac{33}{100}\)
\(E=\dfrac{1}{5.9}+\dfrac{1}{9.13}+\dfrac{1}{13.17}+...+\dfrac{1}{41.45}\)
\(=\dfrac{4}{4.5.9}+\dfrac{4}{4.9.13}+\dfrac{4}{4.13.17}+...+\dfrac{4}{4.41.45}\)
\(=\dfrac{1}{4}\left(\dfrac{4}{5.9}+\dfrac{4}{9.13}+\dfrac{4}{13.17}+...+\dfrac{4}{41.45}\right)\)
\(=\dfrac{1}{4}\left[\left(\dfrac{1}{5}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{13}\right)+\left(\dfrac{1}{13}-\dfrac{1}{17}\right)+...+\left(\dfrac{1}{41}-\dfrac{1}{45}\right)\right]\)
\(=\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)\)
\(=\dfrac{1}{4}.\dfrac{9-1}{45}\)
\(=\dfrac{1}{4}.\dfrac{8}{45}\)
\(=\dfrac{2}{45}\)
\(F=\dfrac{2005.2004-1}{2003.2005+2004}\)
\(=\dfrac{2005\left(2003+1\right)-1}{2003.2005+2004}\)
\(=\dfrac{2005.2003+2005-1}{2003.2005+2004}\)
\(=\dfrac{2003.2005+2004}{2003.2005+2004}\)
\(=1\)
Sửa đề: \(G=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+2018}\right)\)
\(G=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right)...\left(1-\dfrac{1}{1+2+3+...+2018}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left[1-\dfrac{1}{\dfrac{\left(1+2018\right).2018}{2}}\right]\)
\(=\dfrac{3-1}{3}.\dfrac{6-1}{6}.\dfrac{10-1}{10}...\left(\dfrac{\dfrac{2019.2018}{2}-1}{\dfrac{2019.2018}{2}}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\left(\dfrac{\dfrac{2019.2018}{2}-1}{\dfrac{2019.2018}{2}}\right)\)
\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{2.\left(\dfrac{2019.2018}{2}-1\right)}{2019.2018}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\dfrac{2017.2020}{2018.2019}\)
\(=\dfrac{1.2.3...2017}{2.3.4...2018}.\dfrac{4.5.6...2020}{3.4.5...2019}\)
\(=\dfrac{1}{2018}.\dfrac{2020}{3}\)
\(=\dfrac{1010}{3027}\)