`a)`
`(2-x)/x +2=3/x` `(x\ne0)`
`<=>(2-x+2x)/x=3/x`
`=>2-x+2x=3`
`<=>x+2=3`
`<=>x=1` (thoả mãn)
Vậy `S={1}`
`b)`
`x/(x-2)-2/(x+2)=(2x+4)/(x^2-4)` `(x\ne+-2)`
`<=>(x(x+2)-2(x-2))/(x^2-4)=(2x+4)/(x^2+4)`
`=>x(x+2)-2(x-2)=2x+4`
`<=>x^2+2x-2x+4=2x+4`
`<=>x^2+4=2x+4`
`<=>x^2-2x+4-4=0`
`<=>x(x-2)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x-2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=0(TM)\\x=2(KTM)\end{array} \right.\)
Vậy `S={0}`