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\(H=\frac{2\cdot2}{1\cdot5}+\frac{2\cdot2}{5\cdot9}+...+\frac{2\cdot2}{45.49}\)
\(H=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{45\cdot49}\)
\(H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(H=1-\frac{1}{49}\)
\(H=\frac{48}{49}\)
\(H=\frac{2.2}{1.5}+\frac{2.2}{5.9}+\frac{2.2}{9.13}+...+\frac{2.2}{45.49}\)
\(\Rightarrow H=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{45.49}\)
\(\Rightarrow H=\frac{5-1}{1.5}+\frac{9-5}{5.9}+...+\frac{49-45}{45.49}\)
\(\Rightarrow H=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{45}-\frac{1}{49}\)
\(\Rightarrow H=1-\frac{1}{49}=\frac{48}{49}\)
Đặt \(B=\frac{1}{1.2}+\frac{5}{2.7}+\frac{8}{7.15}+\frac{13}{15.28}+\frac{21}{28.49}+\frac{32}{49.81}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}+\frac{1}{28}-\frac{1}{49}+\frac{1}{49}-\frac{1}{81}\)
\(\Rightarrow B=1-\frac{1}{81}\)
\(\Rightarrow B=\frac{80}{81}\)
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
bai nay de thui
nhung bay gio mk ban
luc nao ranh mk lam
cho nha
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Gọi \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=1-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=\frac{24}{25}\)
\(\Leftrightarrow\)\(A=\frac{24}{25}:3\)
\(\Leftrightarrow\)\(A=\frac{24}{25}.\frac{1}{3}\)
\(\Leftrightarrow\)\(A=\frac{8}{25}\)
Vậy \(A=\frac{8}{25}\)
Đặt \(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)
\(\Rightarrow3C=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{22.25}\)
\(\Rightarrow3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Rightarrow3C=1-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow C=\frac{24}{25}:3=\frac{8}{25}\)
Vậy \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)
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
\(A=\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+........+\frac{1}{100.104}\)
\(=\frac{1}{4}.\left(\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+.......+\frac{4}{100.104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+.......+\frac{1}{100}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{5}-\frac{1}{104}\right)\)
\(=\frac{1}{4}.\frac{99}{520}=\frac{99}{2080}\)