$\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}$x+42000 +x+32001 =x+22002 +x+12003
$\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)$⇒(x+42000 +1)+(x+32001 +1)=(x+22002 +1)+(x+12003 +1)
$\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}$⇒x+20042000 +x+20042001 =x+20042002 +x+20042003
$\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0$⇒x+20042000 +x+20042001 −x+20042002 −x+20042003 =0
$\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0$⇒(x+2004)(12000 +12001 −12002 −12003 )=0
$\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\Rightarrow x+2004=0$12000 +12001 −12002 −12003 ≠0⇒x+2004=0
=>x=0-2004
=>x=-2004
vậy x=-2004
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