=>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$
=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{2002} \right )$ - $\left ( \frac{x+1+2003}{2003} \right )$ = $0$
=>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ - $\left ( \frac{x+2004}{2003} \right )$ = 0
=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} -\frac{1}{2002} -\frac{1}{2003}\right )$ =0
=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right )$ $>$$0$
=>$x$+$2004$ = $0$ => $x$$=$ $-2004$