a) ĐK: `x \ne \pm 3`
`5/(x-3)+4/(x+3)=(x-5)/(x^2-9)`
`<=> 5(x+3)+4(x-3)=x-5`
`<=> 5x+15+4x-12=x-5`
`<=> 9x+3=x-5`
`<=>8x=-8`
`<=>x=-1(TM)`
Vậy `S={-1}`
b) ĐK: `x \ne 0 ; x \ne 2`
`(x+2)/(x-2)-1/x=2(x^2-2x)`
`<=> (x+2)/(x-2)-1/x=2/(x(x-2))`
`<=> x(x+2)-(x-2)=2`
`<=>x^2+2x-x+2=2`
`<=>x^2+x=0`
`<=>x(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=0(L)\\x=-1(TM)\end{array} \right.\)
Vậy `S={-1}`
c) ĐK: `x \ne 1 ; x \ne 3`
`2/(x-1)-1/(x-3)=(3x)/(x^2-4x+3)`
`<=> 2/(x-1)-1/(x-3)=(3x)/((x-1)(x-3))`
`<=> 2(x-3)-(x-1)=3x`
`<=>2x-6-x+1=3x`
`<=>-2x=6`
`<=>x=-3 (TM)`
Vậy `S={-3}`