\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-\dfrac{x-1}{2002}-1=0\)  

\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1-\dfrac{x-1}{2002}+1=0\) 

\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\left(\dfrac{x-2003}{2002}\right)=0\)

\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\) \(\Leftrightarrow x=2003\) vì  \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}>0\)Vậy...

Ta có: \(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\)

\(\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}=0\)

\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}+1=0\)

\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\dfrac{x-2003}{2002}=0\)

\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\)

mà \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\ne0\)

nên x-2003=0

hay x=2003

Vậy: S={2003}

Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)

=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)

=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)

=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)

=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)

=> \(x^2-1=0\)

=> \(x^2=1\)

=> \(x=\pm1\)

Vậy phương trình có 2 nghiệm là x = 1, x = -1 .

Thanks bn

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

\(\Leftrightarrow x=-2005\)

\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)

\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\frac{-x}{2003}+1\)

\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)

\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)

\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow2003-x=0\left(\text{ vì }\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\right)\)

<=>x=2003

Vậy S={2003}