`(x+1)^2+(x+1)^2/(x+2)^2=8`
Đặt `x+1=t`
Ta có pt ẩn phụ t là:
`t^2+(t^2)/(t+1)^2=8`
`<=> (t^2(t+1)^2+t^2)/(t+1)^2=8`
`<=> t^2(t^2+2t+1)+t^2=8(t+1)^2`
`<=> t^4+2t^3+t^2+t^2=8(t^2+2t+1)`
`<=> t^4+2t^3+2t^2=8t^2+16t+8`
`<=> t^4+2t^3-6t^2-16t-8=0`
`<=> (t^4-2t^3-2t^2)+(4t^3-8t^2-8t)+(4t^2-8t-8)=0`
`<=> t^2(t^2-2t-2)+4t(t^2-2t-2)+4(t^2-2t-2)=0`
`<=> (t^2-2t-2)(t^2+4t+4)=0`
`<=> (t^2-2t-2)(t+2)^2=0`
`<=>`\(\left[ \begin{array}{l}t^2-2t-2=0\\t+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}(t-1)^2=3\\t=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}t=1+\sqrt[]{3}\\t=1-\sqrt[]{3}\\t=-2\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x+1=1+\sqrt[]{3}\\x+1=1-\sqrt[]{3}\\x+1=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\sqrt[]{3}\\x=-\sqrt[]{3}\\x=-3\end{array} \right.\)
Vậy `S={+-\sqrt{3};-3}`
