Question

Cho biểu thức sau: 1/3 + 1/6 + 1/10 +...+ 2/x(x + 1) = 2007/2009

Tìm x

Answer 1

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\) ( 2/6 = 1/3;2/12=1/6;1/10=2/20;...)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(1-2.\frac{1}{x+1}=\frac{2007}{2009}\)

\(\frac{2}{x+1}=1-\frac{2007}{2009}\)

\(\frac{2}{x+1}=\frac{2}{2009}\)

=> x +1  = 2009

x= 2008

Answer 2

<p>2x[1/6+1/12+1/20+....+1/Xx[x-1]=2007/2008</p><p>2x[1/2x3+1/3x4+1/4x5+....+1/Xx[x-1]=2007/2008</p><p>2x[1/2-1/3+1/3-1/4+1/4-1/5+....+1/[x-1]xX=2007/2008</p><p>1/2-1/x=2007/2008x1/2</p><p>1/2-1/x=2007/4016</p><p>2x[1/2-1/x]=2007/2008</p><p>1/x=1/2-2007/4016 1/x=1/4016.Vay x=4015

Answer 3

\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

\(\Rightarrow2\left(\frac{x-1}{2\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\)

\(\Leftrightarrow x-1=2007\Leftrightarrow x=2008\)

Answer 4

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+........+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)

\(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.......+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{2009}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)

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

=> x + 1 = 2009 => x = 2008

Vậy x = 2008