(x+4/2000 + 1) + (x+3/2001 + 1) = (x+2/2002 + 1) + (x+1/2003 + 1)
x + 2004/2000 + x+2004/2001 = x+2004/2002 + x+2004/2003
(x + 2004) x (1/2000 + 1/2001) = (x + 2004) x (1/2002 + 1/2003)
Với x + 2004 = 0 => x = -2004, ta có: 0 x (1/2000 + 1/2001) = 0 x (1/2002 + 1/2003), đúng
Với x khác 0 => 1/2000 + 1/2001 = 1/2002 + 1/2003, vô lí
Vì 1/2000 + 1/2001 > 1/2002 + 1/2003
vậy x = -2004
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\).Do\(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Leftrightarrow x=-2004\)
