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\(\frac{x+4}{2001}+\frac{x+3}{2002}=\frac{x+2}{2003}+\frac{x+1}{2004}\)
\(\Leftrightarrow\left(\frac{x+4}{2001}+1\right)+\left(\frac{x+3}{2002}+1\right)=\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+1}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}=\frac{x+2005}{2003}+\frac{x+2005}{2004}\)
\(\Leftrightarrow\frac{x+2005}{2001}+\frac{x+2005}{2002}-\frac{x+2005}{2003}-\frac{x+2005}{2004}=0\)
\(\Leftrightarrow\left(x+2005\right).\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)=0\)
Vì \(\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=0-2004=-2004\)
ta có \(2004+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)
\(=\left(1+\frac{2003}{2}\right)+\left(1+\frac{2002}{3}\right)...\left(1+\frac{1}{2004}\right)+1\)
\(=\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}+\frac{2005}{2005}\)
\(=2005\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}}{2005\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)
\(=\frac{1}{2005}\)