= (1/1 -1/2) + (1/2-1/3) + (1/3x1/4)+...+(1/999- 1/1000)

= 1/1- 1/1000

= ...[bn tự tính nhé]

k mk nha, nếu đúng

\(\text{Đề bạn bị sai thì phải ????? Đề đúng phải là }:\)

      \(\frac{1}{1\text{ x }2}+\frac{1}{2\text{ x }3}+\frac{1}{3\text{ x }4}+...+\frac{1}{999\text{ x }1000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\)

\(=1-\frac{1}{1000}\)

\(=\frac{999}{1000}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+........+\frac{1}{999}-\frac{1}{1000}+1\)

=\(\frac{1}{1}-\frac{1}{1000}+1\)

=\(\frac{1999}{1000}\)

     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

= \(1-\frac{1}{1000}+1\)

= \(\frac{999}{1000}+1\)

=\(\frac{1999}{1000}\)

                                                     \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)

                                                   = \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}+1\)

                                                   = \(\frac{1}{1}-\frac{1}{1000}+1\)

                                                   = \(\frac{999}{1000}+1\)

                                                   = \(\frac{999}{1000}+\frac{1000}{1000}\) 

                                                   = \(\frac{1999}{1000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}+1\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1=\frac{1999}{1000}\)

1/2*2+1/2*3+...+1/999*1000+1

=1-1/2+1/2-1/3+1/3-1/4+...+1/999-1/1000+1

=1+1-1/1000

=1999/1000

1/1 - 1/1000 + 1 = 1999/1000

đúng 100% đó bạn , nha

\(\frac{1}{1x2}+\frac{1}{2x3}+.................+\frac{1}{999.1000}+1\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.............+\frac{1}{999}-\frac{1}{1000}+1\)

\(=1-\frac{1}{1000}+1\)

\(=\frac{999}{1000}+1\)

\(=\frac{1999}{1000}\)

\(\frac{999}{1000}+\frac{998}{1000}+......+\frac{1}{1000}\)

\(=\frac{999+998+997+........+1}{1000}\)

\(=\frac{499500}{1000}=\frac{999}{2}\)

\(\frac{999}{1000}+\frac{998}{1000}+.....+\frac{1}{1000}\)

= \(\frac{999+998+.....+1}{1000}\)(cách tính phép tính này rất đơn giản,chỉ việc lấy(999 + 1) x 999 : 2 = ?)

= \(\frac{499500}{1000}=\frac{999}{2}\)

k mình đi! mình trả lời rõ ràng nhất mà