7) 1/x+2 - 1/x-2 = 3x-12/x^2-4 8) 2/x-1 + 1/x+1 = 2x-1/x^2-1

$#DYHUN$

`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}`