\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)

\(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)

\(\frac{1}{x+1}=\frac{1}{2}-\left(\frac{2005}{2007}:2\right)\)

\(\frac{1}{x+1}=\frac{1}{2007}\)

=>x+1=2007

x=2007-1

x=2006

Vậy x=2006

x=2006

$x=2006$

Ta có :\(\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)

          = 2 x \(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\frac{1}{x\left(x+1\right)}\right)=2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)\)

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

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

       Khi đó chỉ cần giải  2 x\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)

Cái bài 2 nhân với 1 là 2/2 nên nhân cả tử cả mẫu với 2 ra 6=2*3

                                                                                              12=3*4

                                                                                               .........

Còn lại tự tính

Nếu ra kết quả đúng thì cho **** nhé

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\frac{2}{5}+...+\frac{2}{x}-\frac{2}{x+1}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{2}-\frac{2}{x+1}=\frac{2005}{2007}\)

\(\Rightarrow1-\frac{2}{x+1}=\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{x+1}=1-\frac{2005}{2007}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2007}\)

\(\Rightarrow x+1=2007\)

\(\Rightarrow x=2006\)

\(\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{2005}{2007}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2005}{4014}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2005}{4014}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{4014}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{4014}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)

\(\Rightarrow x+1=2007\)

\(x=2007-1\)

\(x=2006\)

1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2005/2007

=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2005/2007

=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x*(x+1) = 2005/2007

=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1) = 2005/2007

=> 2(1/2 - 1/x + 1) = 2005/2007

=> 1/2 - 1/x + 1 = 2005/4014

=> 1/x+1 = 1/2007

=> x + 1 = 2007

=> x = 2006

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2005}{2007}\) 

\(\rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2005}{2007}\) 

\(\rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\) 

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2005}{2007}:2\) 

\(\rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2005}{2007}:2\) \(\Rightarrow\frac{1}{x+1}=\frac{1}{2007}\)

\(\Rightarrow x+1=2007\rightarrow x=2006\)

Vậy x = 2006.

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)

<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)

<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0

<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0

<=> x+2011=0

<=> x=-2011

Vậy pt có nghiệm là x=-2011