hmm

Dễ nhỉ?

$c)\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}$

$⇔ \dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1$

$⇔ \dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}$

$⇔ \dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0$

$⇔ (x+2005)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0$

$mà$ $\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right) \neq 0$

$⇒x+2005=0$

$⇔ x=-2005$

Ta có:  x+1/2004 + x+2/2003 = x+3/2002 + x+4/2001

⇔ x+1/2004 + 1 + x+2/2003 + 1 = x+3/2002 + 1 + x+4/2001 + 1

⇔ x+2005/2004 + x+2005/2003 = x+2005/2002+ x+2005/2001

⇔x+2005/2004 + x+2005/2003 - x+2005/2002 - x+2005/2001 = 0

⇔ (x+2005)( 1/2004 + 1/2003 -1/2002 - 1/2001) = 0

Vì 1/2004 + 1/2003 -1/2002 - 1/2001 luôn khác 0

⇒ x+2005 = 0

⇔ x=-2005

Chúc bn học tốt!