$(x+1)^4+(x+3)^4=82$
$⇔(x+2-1)^4+(x+2+1)^4=82$ (1)
Đặt $t=x+2$, pt (1) trở thành:
$⇒(t-1)^4+(t+1)^4=82$
$⇔[(t-1)^2]^2+[(t+1)^2]^2=82$
$⇔(t^2-2t+1)^2+(t^2+2t+1)^2=82$
$⇔(t^2+1-2t)^2+(t^2+1+2t)^2=82$
$⇔(t^2+1)^2-4t(t^2+1)+4t^2+(t^2+1)^2+4t(t^2+1)+4t^2= 82$
$⇔2(t^2+1)^2+8t^2=82$
$⇔2[(t^2+1)^2+4t^2]=82$
$⇔(t^2+1)^2+4t^2=41$
$⇔t^4+2t^2+1+4t^2=41$
$⇔t^4+6t^2-40=0$
$⇔$\(\left[ \begin{array}{l}t^2=4\\t^2=-10(vô-nghiệm)\end{array} \right.\)
$⇔t^2=4$
$⇔$\(\left[ \begin{array}{l}t=2\\t=-2\end{array} \right.\) (Thay $t=x+2$)
$⇔$\(\left[ \begin{array}{l}x+2=2\\x+2=-2\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
