`(x+1)/999+(x+2)/998+(x+3)/997=(x-1)/1001+(x-2)/1002+(x-3)/1003`

`⇔((x+1)/999+1)+((x+2)/998+1)+((x+3)/997+1)=((x-1)/1001+1)+((x-2)/1002+1)+((x-3)/1003+1)`

`⇔(x+1000)/999+(x+1000)/998+(x+1000)/997=(x+1000)/1001+(x+1000)/1002+(x+1000)/1003`

`⇔(x+1000)/999+(x+1000)/998+(x+1000)/997-(x+1000)/1001-(x+1000)/1002-(x+1000)/1003=0`

`⇔(x+1000)(1/999+1/998+1/997-1/1001-1/1002-1/1003)=0`

Mà `1/999+1/998+1/997-1/1001-1/1002-1/1003 > 0`

`=> x+1000=0`

`⇔x=-1000`

Vậy `x=-1000`