lm đúng tui tick cho 2 tick!

x= 56 tick tớ nhé 

bạn giải cách làm dc ko?trình bày dễ hỉu mình vào nit phụ tick cho!

\(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}\)

                                                        \(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}\right)\)

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

\(=2-\frac{2}{x+1}\) mà \(\frac{1}{1}+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)

 => \(2-\frac{2}{x+1}=1\frac{2008}{2010}=>\frac{2}{x+1}=\frac{2}{2010}=>x+1=2010=>x=2009\)

đúng cái nhé 

`Answer:`

\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)

\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Rightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

(*) <=> 1\6 + 1\12 +.. + 1\x.(x+1) = 2009\(2011.2) 
ma 
1\2.3 =1\2-1\3 
1\3.4=1\3-1\4 
............... 
1\x(x+1)= 1\x-1\(x+1) 

cong tung ve ta dc 

Vt= 1\2- 1\(x+1) =2009\(2.2011) 

<=> 2011\(2.2011) -2009\(2.2011) =1\(x+1) 

<=> 1\2011 =1\(x+1) 

=> x=2010

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 1999/2001

nhân 1/2 vào 2 vế ta được vế trái là :

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)

\(\frac{x-1}{2.\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)

\(\frac{x-1}{\left(x+1\right)}=\frac{1999}{2001}\)

suy ra : 2001x - 2001 = 1999x + 1999

2x = 1999 + 2001 = 4000

=> x = 2000

trừ 1 vào mỗi tỉ số,ta đc:

\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}-1=\frac{x-4}{2008}-1\)

\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}=\frac{x-4-2008}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)

\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(mà\frac{1}{2011}<\frac{1}{2010}<\frac{1}{2009}<\frac{1}{2008}\Rightarrow\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

=>x-2012=0

=>x=2012

vậy x=2012

sao cộng mà ko trừ đi

bn -3 đi mà hiệu chỉ có trừ 1 à

\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}\):2

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}\)

\(\Rightarrow\frac{1}{2010}=\frac{1}{x+1}\)

\(\Rightarrow x=2009\)

nha !

Ta có :A=1+\(\frac{2}{6}\)+\(\frac{2}{12}\)+......+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{4018}{2010}\)

\(\Rightarrow\)A=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2008}{2010}\)

\(\Rightarrow\)A=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2008}{2010}\)

\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)

\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)

\(\Rightarrow\)A=\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{502}{1005}\)

\(\Rightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2010}\)\(\Rightarrow\)x+1=2010\(\Rightarrow\)x=2009