Đáp án:
`(x+3)^4+(x+5)^4=16`
Đặt `m= x+4`
`<=> (m -1)^4 + (m+1)^4=16`
`<=> m^4 + 4m^2 + 1 - 4m^3 + 2m^2 - 4m+ m^4 + 4m^2 + 1+ 4m^3 + 2m^2 + 4m= 16`
`<=> 2m^4 + 12m^2 + 2= 16`
`<=> m^4 + 6m^2 + 1= 8`
`<=>m^4 + 6m^2 + 1- 8=0`
`<=> m^4 + 6m^2 - 7=0`
`<=> m^4 + 7m^2 - m^2 - 7=0`
`<=> m^2(m^2 + 7) - (m^2 + 7)=0`
`<=> (m^2 - 1)(m^2 + 7)=0`
`<=>`\(\left[ \begin{array}{l}m^2 - 1=0 \\m^2 + 7=0 \end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}m=±1\\m^2=-7(vô lí)\end{array} \right.\)
Trả lại phép đặt: \(\left[ \begin{array}{l}x+4=1\\x+4=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-3\\x=-5\end{array} \right.\)
Vậy `S={-3; -5}`
