`\qquad x^2+3x+2=24/(x^2-x)` ĐK: `x\ne0;x\ne1`
`<=> (x^2+3x+2)(x^2-x)=24`
`<=> x^4-x^3+3x^3-3x^2+2x^2-2x=24`
`<=> x^4+2x^3-x^2-2x-24=0`
`<=> x^4-2x^3+4x^3-8x^2+7x^2-14x+12x-24=0`
`<=> x^3(x-2)+4x^2(x-2)+7x(x-2)+12(x-2)=0`
`<=> (x-2)(x^3+4x^2+7x+12)=0`
`<=> (x-2)(x^3+3x^2+x^2+3x+4x+12)=0`
`<=> (x-2)[x^2(x+3)+x(x+3)+4(x+3)]=0`
`<=> (x-2)(x+3)(x^2+x+4)=0`
Do `x^2+x+4=(x+1/2)^2+15/4>0` với `AAx`
`=> [(x-2=0),(x+3=0):}`
`<=> [(x=2(\text{tm})),(x=-3(\text{tm})):}`
Vậy `S={2;-3}`