Ta có : 

\(A=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{2}{2}+\frac{2\left(2+1\right)}{2}+\frac{3\left(3+1\right)}{2}+...+\frac{2017\left(2017+1\right)}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+...+\frac{2017.2018}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{\frac{1.2+2.3+3.4+...+2017.2018}{2}}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{1.2+2.3+3.4+...+2017.2018}{2}.\frac{1}{1.2+2.3+3.4+...+2017.2018}\)

\(A=\frac{1}{2}\)

Vậy \(A=\frac{1}{2}\)

Chúc bạn học tốt ~ 

Cảm ơn PMQ nhiều nha cậu cứu mình rồi

ai tích mình mình tích lại cho

tinh tong A= 1.2^2+2.3^2+3.4^2+........+2018.2019^2

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)\)

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

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

\(=2.\frac{2017}{2018}\)

\(=\frac{2017}{1009}\)

quy tử số thành 1 

A = 2.(1/1.2+1/3.2+1/3.4+... + 1/2017.2018) 

A= 2. (1- 1/2018) 

Tính nốt nha

A=\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{2017.2018}\)

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

A=\(2-\frac{2}{2018}\)

A=\(\frac{1017}{509}\)

Tk cho mk mk tk lại

sorry mọi ng nha,toán 6 chứ ko phải toán 5

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2017\cdot2018}\)

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

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

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

b) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)( sửa 91.99 thành 97.99 mới đúng nha )

\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{97}-\frac{2}{99}\right)\)

\(=\frac{1}{2}\left(\frac{2}{3}-\frac{2}{99}\right)\)

\(=\frac{1}{2}.\frac{64}{99}\)

\(=\frac{32}{99}\)

a) 1/1.2 + 1/2.3 + 1/3.4 +...+1/2017.2018

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

= 1 - 1/2018 

= 2017/2018