\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)

\(=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{1999}+\frac{1}{1999}\right)-\frac{1}{2000}\)

\(=\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)

\(=\frac{1}{1}-\frac{1}{2000}\)

\(=\frac{2000}{2000}-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{1999}-\frac{1}{2000}\)

\(=1-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

chịch

Mình ko chép đề nha!

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)

\(=1-\frac{1}{2000}\)

\(=\frac{1999}{2000}\)

Học tốt^^

\(\Leftrightarrow x.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\right)=1\)

\(\Leftrightarrow x.\left(1-\frac{1}{2000}\right)=1\Leftrightarrow x\cdot\frac{1999}{2000}=1\Leftrightarrow x=\frac{2000}{1999}\)

`#3107`

`a)`

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)

\(=1-\dfrac{1}{2000}\)

\(=\dfrac{1999}{2000}\)

`b)`

\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

`c)`

\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)

b) Sửa đề:

 \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}.\dfrac{102}{103}\)

\(=\dfrac{34}{103}\)

c) \(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\left(\dfrac{9}{9}-\dfrac{1}{9}\right)\)

\(=\dfrac{8}{9}-\dfrac{8}{9}\)

\(=0\)

\(#WendyDang\)

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{1999}+\frac{1}{2000}\)

\(S=1-\frac{1}{2000}\)

\(S=\frac{1999}{2000}\)

Đây là bài làm của mk :

S = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/1999 * 2000

=> S = 1 - 1/2 + 1/2 - 1/3 + ... + 1/1999 - 1/2000

=> S = 1 - 1 / 2000 

=> S = 2000/2000 - 1/2000 = 1999/2000

Chúc bn học tốt !

\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1999.2000}\)

\(\Leftrightarrow S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)

\(\Leftrightarrow S=1-\frac{1}{2000}\)

\(\Leftrightarrow S=\frac{1999}{2000}\)

A = 1.2+2.3+3.4+.........................+1999.2000 

=> 3A= 1.2.(3-0) + 2.3.(4-1) + ... + 1999.2000(2001-1998)

=> 3A=1.2.3+2.3.4-1.2.3+...+1999.2000.2001-1998.1999.2000

=> 3A=1999.2000.2001

A=1999.2000.2001:3=2666666000

ban chi can nhan tat co cac do hang voi3 xong sau do ban tinh 

ban chi can nhan cac so hang voi 3 la duoc

đặt là S rồi nhân S và dãy với 3

ta có :