Question

[Nguyễn Demon]

tìm x biết 

1/3 + 1/6 + 1/10 + ......+ 2/x.(x+1)= 2005/2007 ( với x là số tự nhiên khác 0 ) 

helps me 

Answer 1

[Seulgi]

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

Answer 2

[Trường]

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

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

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

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

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

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

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

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

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

Vậy x = 2006.