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HP
17 tháng 3 2016
1/1.2+1/2.3+.....+1/x.(x+1)=2008/2009
=>1/1-1/2+1/2-1/3+.....+1/x-1/x+1=2008/2009
=>1/1+(-1/2+1/2)+(-1/3+1/3)+....+(-1/x+1/x)-1/x+1=2008/2009
=>1/1+0+0+.....+0-1/x+1=2008/2009
=>1-1/x+1=2008/2009
=>1/x+1=1-2008/2009=1/2009
=>x+1=2009
=>x=2008
vậy x=2008
NM
31 tháng 1 2016
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
10 tháng 3 2016
\(\Leftrightarrow\frac{1}{2}+\left(\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}+2.\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3}{10}\)
\(\Leftrightarrow2.\left(\frac{1}{7}-\frac{1}{x+1}\right)=\frac{3}{10}-\frac{1}{2}=-\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{7}-\frac{1}{x+1}=-\frac{1}{5}:2=-\frac{1}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{7}-\left(-\frac{1}{10}\right)=\frac{17}{70}\)
\(\Rightarrow17x+17=70\)
=> không tồn tại n vì n là số tự nhiên
HP
9 tháng 4 2016
\(\frac{7}{12}x+0,75=-2\frac{1}{6}=-\frac{13}{6}\)
\(=>\frac{7}{12}x=-\frac{13}{6}-0,75=-\frac{13}{6}-\frac{3}{4}=-\frac{35}{12}\)
\(=>x=-\frac{35}{12}:\frac{7}{12}=-\frac{35}{12}.\frac{12}{7}=-\frac{35}{7}=-5\)
Vậy x=-5
HP
9 tháng 4 2016
\(-1<\frac{x}{4}<\frac{1}{2}\)
\(<=>-\frac{4}{4}<\frac{x}{4}<\frac{2}{4}\)
<=>-4<x<2
<=>x E {-3;-2;-1;0;1}
Vậy.......................
CN
1
M
26 tháng 4 2016
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\)= \(\frac{1}{4}\)=> x. (\(\frac{1}{2}\)- \(\frac{2}{3}\) + \(\frac{3}{4}\)- \(\frac{5}{6}\)) = \(\frac{1}{4}\)=> x.( \(\frac{6}{12}\)- \(\frac{8}{12}\)+\(\frac{9}{12}\)-\(\frac{10}{12}\))= \(\frac{1}{4}\)=> x. \(\frac{-1}{4}\)=\(\frac{1}{4}\)=> x = \(\frac{1}{4}\): \(\frac{-1}{4}\)=> x = -126 tháng 4 2016
=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
21 tháng 3 2020
Ta có:
\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\...\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\)
=\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}\)
\(=10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{11-10}{10.11}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=10x+A=10x+\frac{10}{11}=11x\)
\(\Rightarrow\frac{10}{11}=11x-10x\)
\(\Rightarrow x=\frac{10}{11}\)
\(\dfrac{x}{2008}+\dfrac{x}{2009}-\dfrac{x}{2007}=1+\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{2}{2007}\)
\(\Rightarrow x = \dfrac{2007.2008.2009+2009.2007-2008.2007-2.2008.2009}{2009.2007+2008.2007-2008.2009}\)
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