Đặt `x^2+2x+2=t`
PT `<=>1/t^2 + 1/(t+1)^2=5/4`
`<=>1/t^2 + 1/(t^2+2t+1)=5/4`
`<=>(2t^2+2t+1)/(t^2(t^2+2t+1))=5/4`
`<=>8t^2+8t+4=5t^4+10t^3+5t^2`
`<=>5t^4+10t^3-3t^2-8t-4=0`
`<=> (t-1)(t+2)(5t^2+5t+2)=0`
`<=>t in {1;-2}`
Vs `t=1 <=>(x+1)^2=0<=>x=-1`
Vs `t=-2 <=>x^2+2x+2=-2` (vô lí)
Vậy `S={-1}`