@py
`\text{Mong bạn cho mình ctlhn ạ!!!}`
`(x+1)/(2009)+(x+3)/(2007)=(x+5)/(2005)+(x+7)/(1993)`
`⇔(x+2010)/(2009)+(x+2010)/(2007)=(x+2010)/(2005)+(x+2010)/(1993)`
`⇔(x+2010)/(2009)+(x+2010)/(2007)-(x+2010)/(2005)-(x+2010)/(1993)=0`
`⇔(x+2010)×(1/(2009)+1/(2007)-1/(2005)-1/(1993))=0`
`⇔x+2010=0`
`⇔x=-2010`