Question

Cứu

Answer 1

`(x+1)/2009+(x+3)/2007=(x+5)/2005+(x+7)/1993`

`<=>(x+1)/2009+1+(x+3)/2007+1=(x+5)/2005+1+(x+7)/1993+1`

`<=>(x+2020)/2009+(x+2020)/2007=(x+2020)/2005+(x+2020)/1993`

`<=>(x+2020)(1/2009+1/2007-1/2005-1/1993)=0`

`<=>x+2020=0`

`<=>x=-2020`

Vậy `S={-2020}`