$#DYHUN$
`1/(x+2)-1/(x-2)=(3x-12)/(x²-4) (1)`
`ĐKXĐ:x`$\neq$`±2`
`(1)⇔(x-2)/((x-2)(x+2))-(x+2)/((x-2)(x+2))=(3x-12)/((x-2)(x+2))`
`⇒x-2-x-2=3x-12`
`⇔x-x-3x=2+2-12`
`⇔-3x=-8`
`⇔x=8/3`
Vậy`S={8/3}`
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`2/(x-1)+1/(x+1)=(2x-1)/(x²-1) (1)`
`ĐKXĐ:x`$\neq$`±1`
`(1)⇔(2(x+1))/((x-1)(x+1))+(x-1)/((x-1)(x+1))=(2x-1)/((x-1)(x+1))`
`⇒2x+2+x-1=2x-1`
`⇔2x-2x+x=1-1-2`
`⇔x=-2`
Vậy `S={2}`
