`(x+3)^4+(x+5)^4=16`
Đặt `y=x+4`
`->(y^2-2y+1)^2+(y^2+2y+1)^2=16`
`->y^4+4y^2+1-4y^3+2y^2-4y+y^4+4y^2+1+4y^3+2y^2+4y=16`
`->2y^4+8y^2+2+4y^2=16`
`->y^4+6y^2+1=8`
`->y^4+6y^2-7=0`
`->(y^2-1)(y^2+7)=0`
`->(y-1)(y+1)(y^2+7)=0`
Vì `y^2+7>0`
`->`\(\left[ \begin{array}{l}y-1=0\\y+1=0\end{array} \right.\)
`->` \(\left[ \begin{array}{l}y=1\\y=-1\end{array} \right.\)
Vì `y=x+4`
`->x=1-4=-3`
`x=1-4=-5`
Vậy `S={-3;-5}`
