Đáp án+Giải thích các bước giải:
`(x+9)/2000+(x+8)/2001=(x+5)/2004+(x+4)/2005`
`\to (x+9)/2000+1+(x+8)/2001+1=(x+5)/2004+1+(x+4)/2005+1`
`\to (x+9+2000)/2000+(x+8+2001)/2001=(x+5+2004)/2004+(x+4+2005)/2005`
`\to (x+2009)/2000+(x+2009)/2001=(x+2009)/2004+(x+2009)/2005`
`\to (x+2009)/2000+(x+2009)/2001-(x+2009)/2004-(x+2009)/2005=0`
`\to (x+2009)(1/2000+1/2001-1/2004-1/2005)=0`
`\to x+2009=0` (vì `1/2000+1/2001-1/2004-1/2005\ne0`)
`\to x=-2009`
Vậy `x=-2009`
