Giải thích các bước giải:
`x (3-x)(x+1) (x+(3-x)/(x+1))=2`
`<=>(x(3-x))/(x+1) . (x(x+1)+(3-x))/(x+1)=2`
`<=>(3x-x^2)/(x+1) . (x^2+x+3-x)/(x+1)=2`
`<=>(3x-x^2)/(x+1) . (x^2+3)/(x+1)=2`
`<=>((3x-x^2)(x^2+3))/(x+1)^2=2`
`<=>3x^3+9x-x^4-3x^2=2(x+1)^2`
`<=>-x^4+3x^3-3x^2+9x=2(x^2+2x+1)`
`<=>-x^4+3x^3-3x^2+9x=2x^2+4x+2`
`<=>-x^4+3x^3-5x^2+5x-2=0`
`<=>x^4-3x^3+5x^2-5x+2=0`
`<=>(x^4-x^3)-(2x^3-2x^2)+(3x^2-3x)-(2x-2)=0`
`<=>(x-1)(x^3-2x^2+3x-2)=0`
`<=>(x-1)[(x^3-x^2)-(x^2-x)+(2x-2)]=0`
`<=>(x-1)(x-1)(x^2-x+2)=0`
`<=>(x-1)^2(x^2-x+2)=0`
Mà `x^2-x+2>0AAx=>(x-1)^2=0`
`<=>x-1=0<=>x=1`
Vậy `S={1}.`
