`a)`
`(3(x-1))/4+1>(x+2)/3`
`<=> (9(x-1)+12)/12>(4(x+2))/12`
`<=> 9(x-1)+12>4x+8`
`<=> 9x-9+12>4x+8`
`<=> 9x-4x>8-12+9`
`<=> 5x>5`
`<=> x>1`
Vậy `S={x|x>1}`
`b)` ĐKXĐ : `x ne +-2`
`(x-2)/(x+2)-3/(x-2)=(2(x-11))/(x^2-4)`
`<=> ((x-2)^2-3(x+2))/(x^2-4)=(2x-22)/(x^2-4)`
`=> (x-2)^2-3(x+2)=2x-22`
`<=> x^2-4x+4-3x-6=2x-22`
`<=> x^2-7x-2x+4-6+22=0`
`<=> x^2-9x+20=0`
`<=> x^2-4x-5x+20=0`
`<=> x(x-4)-5(x-4)=0`
`<=> (x-5)(x-4)=0`
`<=> [(x=5 \ \ \text{(tm)}),(x=4 \ \ \text{(tm)}):}`
Vậy `S={4;5}`
