Tìm X biết : X+3/2008-X+3/2008-X+3/2010-X+3/2009
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\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\left(x+3\right)\cdot\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)
=> x + 3 = 0
x = -3
\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)
\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)
\(\Leftrightarrow x+3=0\)
hay x=-3
\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
\(\Leftrightarrow x=2012\)
\(\frac{x-2012}{2}+\frac{x-2008}{3}+\frac{x-2002}{4}+\frac{x-1994}{5}=10\)
\(\Leftrightarrow\frac{x-2012}{2}-1+\frac{x-2008}{3}-2+\frac{x-2002}{4}-3+\frac{x-1994}{5}-4=0\)
\(\Leftrightarrow\frac{x-2014}{2}+\frac{x-2014}{3}+\frac{x-2014}{4}+\frac{x-2014}{5}=0\)
<=> x = 2014(vì 1/2 + 1/3 + 1/4 + 1/5 khác 0)
Ta có :\(A=3+3^2+3^3+...+3^{2008}\)(1)
\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)(2)
Lấy (2) trừ đi 1 ta có :
\(\Rightarrow2A=3^{2009}-3\)
Ta lại có :
\(2A+3=3^x\)
\(\Rightarrow3^{2009}=3^x\)
\(\Rightarrow x=2009\)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2008}{2010}.\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2008}{2010}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\frac{1}{x+1}=\frac{1}{2010}\)
=> x + 1 = 2010
=> x = 2009
Ta có : \(\frac{2}{2\times3}+\frac{2}{3\times4}+....+\frac{2}{x\times\left(x+1\right)}=\frac{2008}{2010}\)
\(\Rightarrow2\times\left(\frac{1}{2\times3}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{1004}{1005}\)
\(\Rightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1004}{1005}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1004}{1005}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{502}{1005}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{502}{1005}=\frac{1}{2010}\)
\(\Rightarrow x+1=2010\)
\(\Rightarrow x=2010-1=2009\)
minh tinh ra x = -3
viết thế này bố thằng nào hiểu được