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a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-..........-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2018}{2018}-\frac{1}{2018}=\frac{2017}{2018}\)
b) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+..........+\frac{2}{2017.2018}+\frac{2}{2018.2019}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}+\frac{1}{2018.2019}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.........-\frac{1}{2018}+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=2\left(1-\frac{1}{2019}\right)\)
\(=2\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)
\(=2.\frac{2018}{2019}\)
\(=\frac{4036}{2019}\)
Phần c tương tự nha
a) \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + .......+ \(\frac{1}{2017.2018}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + .......+ \(\frac{1}{2017}\) - \(\frac{1}{2018}\)
= 1 - \(\frac{1}{2018}\) = \(\frac{2017}{2018}\)
câu a) mik sửa đề một tí ko biết có đúng ko
câu b , c tương tự nhưng cần lấy tử ra chung
a)\(\frac{2013}{2015}< \frac{2014}{2016}\)
b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)
\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)
bai nay de thui
nhung bay gio mk ban
luc nao ranh mk lam
cho nha
minhpham@gmail.com
Mik c thôi
A = 1 / 1008 + 1 / 2013 - 1 / 2016 x 2017
A = 1 / 1008 + 1 / 2013 - 1 / 2016 x 1 / 2017
B = 1 / 2014 + 1 / 2016 + 1 / 2017 + 1 / 2014 x 2016
B = 1 / 2014 + 1 / 2016 + 1 / 2017 + 1 / 2014 x 1 / 2016