Gọi \(A=\frac{1005}{2011}\)

A=1/3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A=1/1.3 + 1/3.5 + 1/5.7 +...............+1/x.(x+2)

A . 2=2/1.3 + 2/3.5 + 2/5.7 +......................+2/x.(x+2)

A . 2=1/1-1/3+1/3-1/5+1/5-1/7+..............+1/x-1/x+2

A . 2=1/1+(1/3-1/3)+(1/5-1/5)+..............+(1/x-1/x)-1/x+2

A . 2=1/1-1/x+2

Suy gia:1005/2011 . 2=1/1-1/x+2

             2010/2011    =1/1-1/x+2

             1/x+2           =1/1-2010/2011

              1/x+2          =1/2011

Suy gia:x+2=2011

            x    =2011-2

            x    =2009

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x.\left(x+2\right)}=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)

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

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

\(\frac{1}{x+2}=\frac{1}{2011}\)

\(\Rightarrow x+2=2011\)

\(x=2009\)

Đặt biểu thứ là A

2A=2/1.3+2/2.5+...+2/x.x+2

2A=1-1/3+1/3-1/5+.......+1/x-1/x+2

2A=1-1/x+2

x=2009 nha ban

chuc ban hoc gioi

tk cho minh nha

\(\dfrac{1}{1.3}\)+ \(\dfrac{1}{3.5}\)+ \(\dfrac{1}{5.7}\)+....+\(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{1005}{2011}\)

1- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{7}\)+....+\(\dfrac{1}{x}\)- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)

1- \(\dfrac{1}{x+1}\)= \(\dfrac{1005}{2011}\)

\(\dfrac{1}{x+1}\)= 1- \(\dfrac{1005}{2011}\)

\(\dfrac{1}{x+1}\)= \(\dfrac{1006}{2011}\)

=> x +1= 2011

=> x= 2011-1

=> x=2010

Bài này mk lm đại nha bn ! Cs j sai mong bn bỏ qua .

ko biết

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)

a, 2/3 của -420 là :

-420 x 2/3 = -280

Số cần tìm là :

-280 x 5/8 = -175

Vậy số cần tìm là -175

b, 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/x ( x + 2 ) = 1005 / 2011

1/2 x ( 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/ ( x ( x + 2 ) = 1005 / 2011

1/2 x ( 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/ x + 2 ) = 1005 / 2011

1/2 x ( 1 - 1/ x + 2 ) = 1005 / 2011

1 - 1 / x + 2 = 1005 / 2011 : 1/2 

1 - 1 / x + 2 = 2010 / 2011

x + 2 / x + 2 - 1 / x + 2 = 2010 / 2011

x + 2 - 1 / x + 2 = 2010 / 2011

x + 1 / x + 2 = 2010 / 2011

+> x + 1 = 2010 

x = 2010 - 1 

x = 2009

+> x + 2 = 2011 

x = 2011 - 2 

x = 2009 

Vậy x = 2009 

Tk nha Đúng đó !!

= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2 . (1/1 - 1/2011)

= 1/2 . 2010 / 2011

= 1005/2011

= 1 - 1/2011

= 2010 / 2011

   1/2 ( 2/1.3 + 2/3.5 +...+ 2 /2009.2011)

= 1/2 ( 1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)

= 1/2( 1/1 - 1/2011)

= 1/2 . 2010 / 2011

=1005/2011

\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)

\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)

\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)

\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)

\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)

\(\Rightarrow A=\frac{506521}{2013}\)