1/2+1/6+1/12+1/20+...+1/x.[x+1]=2008/2009
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LE
1
26 tháng 4 2016
=>1/1.2+1/2.3=1/3.4+........+1/x.(x+1)=2008/2009
=>1-1/2+1/2-1/3+.....+1/x-1/x+1=1-1/2009
=>1-1/x+1=1-1/2009
=>-1/x=-1/2009
=>1/x=1/2009
=>x=2009
Nhớ k cho mình nha
NQ
1
S
24 tháng 4 2019
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{1}{x+1}=1-\frac{2008}{2009}=\frac{1}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
PH
2
HP
25 tháng 4 2016
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(=>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(=>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>1-\frac{1}{x+1}=\frac{2008}{2009}=>\frac{1}{x+1}=1-\frac{2008}{2009}=\frac{1}{2009}\)
=>x+1=2009
=>x=2008
Vậy x=2008
25 tháng 4 2016
1/2+1/6+1/12+...+1/x*(x+1)=2008/2009
1/1*2+1/2*3+1/3*4+...+1/x*(x+1)=2008/2009
1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/(x+1)=2008/2009
1-1/x+1)=2008/2009
1/x+1=1-2008/2009
1/x+1=1/2009
nên x+1=2009
x=2009-1
x=2008 (tick nha
)
4 tháng 5 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009
}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1-1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(2009x=2008\left(x+1\right)\)
\(2009x=2008x+2008\)
\(2009x-2008x=2008\)
\(x=2008\)
Vậy x=2008
DT
3
17 tháng 4 2016
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow x+1=2009\)
\(\Leftrightarrow x=2008\)
Vậy x = 2008
17 tháng 4 2016
\(=>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x.\left(x+1\right)}=\frac{2008}{2009}\)
\(=>1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>\frac{x}{x+1}=\frac{2008}{2009}=>x=2008\)
NN
5
DT
13 tháng 5 2017
a) =\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+....+\(\frac{1}{10\cdot11}\)=\(\frac{1}{2}\)-\(\frac{1}{11}\)=\(\frac{9}{22}\)
b)\(\frac{2008}{2009}\)=1 - \(\frac{1}{2009}\); \(\frac{2007}{2008}\)=1 - \(\frac{1}{2008}\)Do \(\frac{1}{2009}\)<\(\frac{1}{2008}\)nen 1 - \(\frac{1}{2009}\)>1 - \(\frac{1}{2008}\)
=> 2008/2009>2007/2008
10 tháng 11 2016
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
10 tháng 11 2016
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
=>1- 1/2 + 1/2 - 1/3+.....+1/x - 1/(x+1) = 2008/2009
=>1 - 1/(x+1) = 2008/2009
=>1 - 1/(x+1) =1-1/1009
=>1/(x+1)=1/2009
=>x+1=2009
=>x=2008.Vậy x=2008