$(x+3)^{4}$ + $(x+5)^{4}$ = 2
Đặt x +4= t, ta có:
<=> $(t-1)^{4}$ + $(t+1)^{4}$ = 2
<=> (t-1)(t³-3t²+3t-1) + $t^{4}$ + 4t³ + 6t² + 4t + 1 =2
<=> $t^{4}$ - 3t³ + 3t² -t -t³+3t²-3t+1+$t^{4}$ +4t³+6t²+4t+1-2=0
<=> $2t^{4}$ + 12t² = 0
<=> 2t²(t²+6)=0
<=> \(\left[ \begin{array}{l}2t^2=0\\t^2+6=0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}t=0\\t^2=-6\end{array} \right.\) (VL)
<=> x+4=0
<=> x=-4
