\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2}{5}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{10}\)

=> x+1=10

=>x=9

cacs bạn 

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}\)

= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)

\(=\frac{1}{2}-\frac{1}{5}=\frac{3}{10}\)

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 

= 1 - 1/5 

=4/5

neu dung thi

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)

Đặt A , ta có :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{999\times1000}+1\)

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

\(A=2-\frac{1}{1000}\)

\(A=\frac{2000}{1000}-\frac{1}{1000}\)

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

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}+1=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}+1\)

\(A=1-\frac{1}{1000}+1=\frac{999}{1000}+1=\frac{1999}{1000}\)

Vậy \(A=\frac{1999}{1000}\)

Giải:

Ta có:  1/1x2+1/2x3+1/3x4+...+1/999x1000+1

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

= 1 - 1/1000 + 1

= 2 - 1/1000

= 1999/1000

Ai tích mk mk sẽ tích lại 

Ko đc Coppy

CHỉ đc viết thui nha mk cho 1 tích  

1999 / 1000 nha Hoàng Tử Ban Mai

\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2010.2011}\)

\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)

\(S=1-\frac{1}{2011}\)

\(S=\frac{2010}{2011}\)