$(x+1)^4 + (x-3)^4 = 82$ $(1)$
Đặt $x-1=t$
$\to x = t+1$
$pt(1):(t+1+1)^4+(t+1-3)^4=82$
$\to (t+1+1)^4+(t+1-3)^4=82$
$\to (t+2)^4+(t-2)^4=82$
$\to t^4+8t^3+24t^2+32t+16+t^4-8t^3+24t^2-32t+16=82$
$\to 2t^4+48t^2+32=82$
$\to t^4+24t^2+16=41$
$\to t^4+24t^2-25=0$
Đặt $t^2=u(u \geq 0)$
$\to t^4=u^2$
$\to u^2+24u-25=0$
$\to (u-1)(u+25)=0$
$\to$ \(\left[ \begin{array}{l}u=1\\u=-25\end{array} \right.\) (loại $u=-25$ vì $u \geq 0)$
$\to (t^2-1)(t^2+25)=0$
$\to$\(\left[ \begin{array}{l}t=±1\\t^2=-25\end{array} \right.\) (loại $t^2=-25$, vì $t^2 \geq 0∀t)$
Hay: \(\left[ \begin{array}{l}x-1=-1\\x-1=1\end{array} \right.\)
$\to$ \(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy: ...
