Đáp án:Điều kiện:`x ne 0`.
`a)x^2+x+1/x+1/x^2-4=0`
`<=>x^2+1/x^2+(x+1/x)-4=0`
`<=>x^2+2+1/x^2+(x+1/x)-6=0`
`<=>(x+1/x)^2+(x+1/x)-6=0`
`<=>(x+1/x)^2+3(x+1/x)-2(x+1/x)-6=0`
`<=>(x+1/x)(x+1/x+3)-2(x+1/x+3)=0`
`<=>(x+1/x+3)(x+1/x-2)=0`
`**x+1/x+3=0`
`<=>x^2+1+3x=0`
Ta có:`Delta=9-4=5`
`<=>` \(\left[ \begin{array}{l}x=\dfrac{-b+\sqrt{\Delta}}{2a}\\x=\dfrac{-b-\sqrt{\Delta}}{2a}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{3-\sqrt5}{2}\\x=\dfrac{3+\sqrt5}{2}\end{array} \right.\)
`**x+1/x-2=0`
`<=>x^2+1-2x=0`
`<=>(x-1)^2=0`
`<=>x=1`
Vậy `S={1,(3+sqrt5)/2,(3-sqrt5)/2}`.