`a)` `(x+1)/9 + (x+2)/8 = (x+3)/7 + (x+4)/6`

`(x+1)/9 + 1 + (x+2)/8 + 1 = (x+3)/7 + 1 + (x+4)/6 + 1`

`(x+10)/9 + (x+10)/8 - (x+10)/7 - (x+10)/6 = 0`

`(x+10)(1/9 + 1/8 - 1/7 - 1/6) = 0`

Vì `1/9 + 1/8 - 1/7 - 1/6` $\neq$ `0`

`=> x+10 = 0`

`x=-10`

Vậy `x=-10`

`b)` `(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 = 0`

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

Vì `1/2004 + 1/2003 - 1/2002 - 1/2001` $\neq$ `0`

`=> x-2005 = 0`

`x=2005`

Vậy `x=2005`

`c)` `(x-45)/55 + (x-47)/53 = (x-55)/45 + (x-53)/47`

`(x-45)/55 - 1 + (x-47)/53 - 1 = (x-55)/45 - 1 + (x-53)/47 - 1`

`(x-100)/55 + (x-100)/53 - (x-100)/45 - (x-100)/47 = 0`

`(x-100)(1/55 + 1/53 - 1/45 - 1/47) =0`

Vì `(1/55 + 1/53 - 1/45 - 1/47)`$\neq$ `0`

`=> x-100 = 0`

`x=100`

Vậy `x=100`