a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 2006 ) + 2007 = 2007

\(\Rightarrow\)( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 2006 + 2007 ) = 2007

\(\Rightarrow\)2007x + 2015028 = 2007

\(\Rightarrow\)2007x = 2007 - 2015028 = -2013021

\(\Rightarrow\)x = ( -2013021 ) : 2007 = -1003

Vậy x = -1003

b) 2000 + ( 199 + x ) + ( 198 + x ) + ... + ( x + 1 ) + x = 200

\(\Rightarrow\)( x + x + x + ... + x + x ) + ( 1 + 2 + ... + 198 + 199 + 2000 ) = 200

\(\Rightarrow\)200x + 2001000 = 200

\(\Rightarrow\)200x = 200 - 2001000 = -2000800

\(\Rightarrow\)x = ( -2000800 ) : 200 = -10004

Vậy x = -10004 

a, x + ( x + 1 ) + ( x + 2 ) + ..... + ( x + 2006) + 2007 = 2007

x. 2007 + ( 1 + 2 + ..... + 2006 ) = 2007 - 2007

x. 2007 + 2013021 = 0

x. 2007                 = 0 - 2013021

x.2007                  = - 2013021

    x                       = ( - 2013021 ) : 2007

   x                        = - 1003

\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)

\(\Leftrightarrow x=2010\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)

\(\Rightarrow x-2010=0\Rightarrow x=2010\)

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)

<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)

<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0

<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0

<=> x+2011=0

<=> x=-2011

Vậy pt có nghiệm là x=-2011

Tag thầy Lâm không :)???

(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010 
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1] 
=(x-2011)/2007+(x-2011)/2008=(x-2011)/... 
=(x-2011)(1/2007+1/2008-1/2009-1/2010)... 
suy ra x=2010