2004x−1+ 2003x−2− 2002x−3= 2001x−4
⇒2004x−1+ 2003x−2− $\dfrac{x-3}{2002}-\dfrac{x-4}{2001}=$`0`
⇒`(`2004x−1−1 `)+(`2003x−2−1`)-(` $\dfrac{x-3}{2002}-1)-(\dfrac{x-4}{2001}-1$`)=0`
⇒2004x−2005+ 2003x−2005− 2002x−2005− 2001x−2005=`0`
⇒`(x-2005)(1/2004+1/2003-1/2002-1/2001)=0`
Mà `1/2004+1/2003-1/2002-1/2001`̸= `0`
⇒`x-2005=0`
⇒`x=2005`
