h) `x^2+\frac{x^2}{(x+1)^2}=8` (ĐKXĐ: `x \ne -1`)
`⇔\frac{x^2(x+1)^2}{(x+1)^2}+\frac{x^2}{(x+1)^2}=\frac{8(x+1)^2}{(x+1)^2}`
`⇒x^2(x^2+2x+1)+x^2=8(x^2+2x+1)`
`⇔x^4+2x^3+x^2+x^2-8x^2-16x-8=0`
`⇔x^4+2x^3-6x^2-16x-8=0`
`⇔x^4+2x^3-6x^2-12x-4x-8=0`
`⇔x^3(x+2)-6x(x+2)-4(x+2)=0`
`⇔(x+2)(x^3-6x-4)=0`
`⇔(x+2)(x^3+2x^2-2x^2-4x-2x-4)=0`
`⇔(x+2)[x^2(x+2)-2x(x+2)-2(x+2)]=0`
`⇔(x+2)(x+2)(x^2-2x-2)=0`
`⇔(x+2)^2(x^2-2x-2)=0`
`⇔(x+2)^2(x^2-2x+1-3)=0`
`⇔(x+2)^2[(x-1)^2-3]=0`
`⇔(x+2)^2(x-1-\sqrt{3})(x-1+\sqrt{3})=0`
`⇔`\(\left[ \begin{array}{l}(x+2)^2=0\\x-1-\sqrt{3}=0\\x-1+\sqrt{3}=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\ (\text{thỏa mãn điều kiện})\\x=\sqrt{3}+1\ (\text{thỏa mãn điều kiện})\\x=-\sqrt{3}+1\ (\text{thỏa mãn điều kiện})\end{array} \right.\)
Vậy `S={-2;\sqrt{3}+1;-\sqrt{3}+1}`