Tìm x :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
24 tháng 4 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2}{4018}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2008\)
R
24 tháng 4 2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=>\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)(nhân cả hai vế với \(\frac{1}{2}\))
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)= \(\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\frac{1}{x+1}\)=\(\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x+1}=\frac{1}{2009}\)
x+1=2009
x=2009-1=2008
Vậy x bằng 2008
28 tháng 4 2017
Ta có :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2007}{2009}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Rightarrow x=2009-1\)
\(\Rightarrow x=2008\)
28 tháng 4 2017
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\)\(2.\)(\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x\left(x+1\right)}\)) \(=\frac{2007}{2009}\)
\(\Rightarrow\)(\(\frac{1}{2}-\frac{1}{x+1}\))\(=\frac{2007}{2009}:2\)
\(\Rightarrow\frac{-1}{x+1}=\frac{2007}{4018}-\frac{1}{2}\)
\(\Rightarrow\frac{-1}{x+1}=\frac{-1}{4018}\Rightarrow x+1=4018\Rightarrow x=4017\)
5 tháng 4 2018
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{2007}{2009}\)
=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(1-\frac{2}{x+1}=\frac{2007}{2009}\)
=> \(\frac{2}{x+1}=\frac{2}{2009}\) => x + 1 = 2009 => x = 2008
8 tháng 8 2015
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)
\(2\cdot\left(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}\right)=\frac{2007}{2009}\)
\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x-1\right)}=\frac{2007}{2009}:2\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(x-1\right)x}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x-1}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2009}\)
=> x = 2009
8 tháng 8 2015
\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) trong đó 3 = x ; 2 = x - 1
\(\frac{1}{\left(x-1\right)x}=\frac{1}{x-1}-\frac{1}{x}\)
4 tháng 8 2015
ĐẶt A = \(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{30}\cdot\cdot\cdot+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{\left(x-1\right)x}-\frac{2007}{2009}\)
A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}-\frac{2007}{2009}\)
A \(=\frac{1}{2}-\frac{1}{x}-\frac{2007}{2009}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow2009=x+1\)
\(\Rightarrow x=2008\)