`1)(x+1)/2020+(x+2)/2019=(x+3)/2018+(x+4)/2017`
`⇔(x+1)/2020+1+(x+2)/2019+1=(x+3)/2018+1+(x+4)/2017+1`
`⇔(x+1+2020)/2020+(x+2+2019)/2019=(x+3+2018)/2018+(x+4+2017)/2017`
`⇔(x+2021)/2020+(x+2021)/2019=(x+2021)/2018+(x+2021)/2017`
`⇔(x+2021)(1/2020+1/2019-1/2018-1/20017)=0`
`⇔x+2021=0`
`⇔x=-2021`
`⇒` Vậy `S={-2021}`
2)`(59-x)/41+(57-x)/43+(55-x)/45+(53-x)/47=0`
`⇔(59-x)/41+1+(57-x)/43+1+(55-x)/45+1++(53-x)/47+1=0`
`⇔(59-x+41)/41+(57-x+43)/43+(55-x+45)/45+(53-x+47)/47=0`
`⇔(100-x)/41+(100-x)/43+(100-x)/45+(100-x)/47=0`
`⇔(100-x)(1/41+1/43+1/45+1/47)=0`
`⇔100-x=0`
`⇔x=-100`
`⇒`Vậy `S={-100}`
`3)(x-3)/(x+3)-(x+3)/(x-3)=(4x^2)/(9-x^2)`
`⇔(x-3)/(x+3)-(x+3)/(x-3)=(-4x^2)/(x^2-3^2)`
`ĐKXĐ:(x+3)(x-3)\ne0`
$x\neq±3$
`⇔(x-3)^2/((x+3)(x-3)) - (x+3)^2/((x+3)(x-3))=(-4x^2)/((x+3)(x-3))`
`⇔x^2-6x+9-(x^2+6x+9)=-4x^2`
`⇔x^2-6x+9-x^2-6x-9+4x^2=0`
`⇔-12x+4x^2=0`
`⇔4x(-3+x)=0`
\(\left[ \begin{array}{l}4x=0\\-3+x=0\end{array} \right.\)
`⇒x=0(nhận)`
`⇒x=3(loại)`
`⇒` Vậy `S={3}`
