A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)

A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)

A=1-\(\frac{1}{2018}\)

A=\(\frac{2017}{2018}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)

\(A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)

Đến đây bình thường ta nhóm 2 số vào với nhau nhưng ở đây có lẻ số hạng nên không nhóm được => đề sai

Lời giải:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2017.2018}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\)

\(3028B=\frac{1010+2018}{1010.2018}+\frac{1011+2017}{1011.2017}+..+\frac{2018+1010}{2018.1010}\)

\(=(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+...+\frac{1}{1010})+(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018})\)

\(=2A\)

\(\Rightarrow \frac{A}{B}=1514\in \mathbb{Z}\)

Lời giải:

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2017.2018}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)

\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\)

\(3028B=\frac{1010+2018}{1010.2018}+\frac{1011+2017}{1011.2017}+..+\frac{2018+1010}{2018.1010}\)

\(=(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+...+\frac{1}{1010})+(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018})\)

\(=2A\)

\(\Rightarrow \frac{A}{B}=1514\in \mathbb{Z}\)

A=\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)

A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)

A=1-\(\frac{1}{2018}\)

A=\(\frac{2018}{2018}\)-\(\frac{1}{2018}\)

A=\(\frac{2017}{2018}\)

Vậy A=\(\frac{2017}{2018}\)

Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

giúp mk vs mn ơi. mình cần gấp chiều mai nộp òi

Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\) 

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)

\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)

\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)

Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)

\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)

\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)

\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)

\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+.....+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.....+\frac{1}{50}^{ĐPCM}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )

Đặt A = ( 1/1.2 + 1/3.4  + ... + 1/2017.2018)

= 1 - 1/2 + 1/3 - 1/4  + ... + 1/2017  - 1/2018

= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )

= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )

= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )

= 1/1010 + 1/1011 + ... + 1/2018

=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018

=> S = 1/2018

Vậy S = 1/2018

thanks bạn nhiều