1.
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{2017\cdot1018}\)
b) \(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+....+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\)
c) \(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+......+\frac{4}{1999\cdot2000}\)
********Lưu ý :
1) Các dấu chấm ở phân số là dấu nhân ạ..!!!
2) Trình bày rõ ràng giúp mk với ạ..!!!
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)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b)nhóm 2 ra ngoài rồi làm như câu a
c)nhóm 4 ra rồi làm như câu a
Trả lời
a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
b) Đặt B=\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\)
\(\Leftrightarrow\frac{1}{2}B=\frac{1}{2}\left(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2017\cdot2018}+\frac{2}{2018\cdot2019}\right)\)
\(\Leftrightarrow\frac{1}{2}B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)
\(\Leftrightarrow\frac{1}{2}B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(\Leftrightarrow\frac{1}{2}B=1-\frac{1}{2019}=\frac{2018}{2019}\)
\(\Leftrightarrow B=\frac{2018}{2019}:\frac{1}{2}=\frac{2018}{2019}\cdot2=\frac{4036}{2019}\)
c) Đặt C=\(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)
\(\Leftrightarrow\frac{1}{4}C=\frac{1}{4}\left(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{1999\cdot2000}\right)\)
\(\Leftrightarrow\frac{1}{4}C=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)
\(\Leftrightarrow\frac{1}{4}C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(\Leftrightarrow\frac{1}{4}C=1-\frac{1}{2000}\)
\(\Leftrightarrow\frac{1}{4}C=\frac{1999}{2000}\)
\(\Leftrightarrow C=\frac{1999}{2000}:\frac{1}{4}=\frac{1999}{2000}\cdot4=\frac{1999}{500}\)
