Đáp án:
Giải thích các bước giải:
$\frac{x+7}{2000}$ + $\frac{x+6}{2001}$ = $\frac{x+5}{2002}$ + $\frac{x+4}{2003}$
(=) $\frac{x+7}{2000}$ +1 + $\frac{x+6}{2001}$+1 = $\frac{x+5}{2002}$ +1+ $\frac{x+4}{2003}$+1
(=)$\frac{x+2007}{2000}$ + $\frac{x+2007}{2001}$ = $\frac{x+2007}{2002}$ + $\frac{x+2007}{2003}$
(=)$\frac{x+2007}{2000}$+$\frac{x+2007}{2001}$ - $\frac{x+2007}{2002}$-$\frac{x+2007}{2003}$=0
(=) ( x+2007) ( $\frac{1}{2000}$+$\frac{1}{2001}$- $\frac{1}{2002}$- $\frac{1}{2003}$)=0
(=) x+2007 =0
(=) x =-2007