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

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

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

\(=\dfrac{1999}{1000}\).

dễ lắm nha

các bạn ơi giúp nhanh nha mình đang cần rất gấp

\(B=\frac{\frac{2016}{1000}+\frac{2016}{999}+...+\frac{2016}{501}}{\frac{-1}{1.2}+\frac{-1}{3.4}+...+\frac{-1}{999.1000}}=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{999.1000}\right)}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{1000}\right)}\)

\(=\frac{2016\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1000}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{999}+\frac{1}{1000}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{500}\right)\right]}\)

\(=\frac{2016.\left(\frac{1}{1000}+\frac{1}{999}+...+\frac{1}{501}\right)}{-\left(\frac{1}{501}+\frac{1}{502}+\frac{1}{503}+....+\frac{1}{999}+\frac{1}{1000}\right)}=\frac{2016}{-1}=-2016\)

Vậy B = - 2016

Bạn Xyz cho mik hỏi ở phần mẫu số tại sao lại có -2*(1/2+1/4+...+1/1000) vậy? Nó ở đâu ra thế?

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

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

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

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

1/1*2+1/2*3+1/3*4+...+1/999*1000

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

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

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

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

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

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

=1/1-1/2+1/2-1/3+,,,,+1/999-1/1000+1

=1-1/1000+1

=1+1-1/1000

=2-1/1000

=1999/1000

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

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

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

\(\Rightarrow A=1-\frac{1}{1000}=\frac{999}{1000}\)

Thay vào ta có : \(\frac{999}{1000}+1=\frac{1999}{1000}\)

Vậy ...

1 / 1x2 + 1 / 2x3 + 1/ 3x4 +....+ 1/999x1000 + 1

= ( 1 / 1 x2 + 1/ 2 x3 + 1/ 3x4 + ....+ 1/999x 1000 ) + 1

= ( 1 x1 / 1 x2 + 1x2 / 2x3 + 1x 3 / 3x4 + ..... 1x 999 / 999x 1000 ) + 1

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

= ( 1 - 1 / 1000 ) + 1

= 999 / 1000 + 1

= 1999 / 1000

Tk tớ nha