Nhận thấy \(x=2;3\) là 2 nghiệm của pt
- Nếu \(x>3\Rightarrow\left\{{}\begin{matrix}\left(x-3\right)^4>0\\\left(x-2\right)^4>1\end{matrix}\right.\) \(\Rightarrow\left(x-2\right)^4+\left(x-3\right)^4>1\Rightarrow ptvn\)
- Nếu \(x< 2\Rightarrow x-3< -1\Rightarrow\left(x-3\right)^4>1\)
\(\Rightarrow\left(x-2\right)^4+\left(x-3\right)^4>1\Rightarrow ptvn\)
- Nếu \(2< x< 3\Rightarrow\left(x-2\right)^4+\left(x-3\right)^4=1\)
\(\Leftrightarrow\left(x-2\right)^4+\left(3-x\right)^4=1\)
\(\left\{{}\begin{matrix}0< x-2< 1\\0< 3-x< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^4< x-2\\\left(3-x\right)^4< 3-x\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^4+\left(3-x\right)^4< x-2+3-x=1\Rightarrow ptvn\)
Vậy pt chỉ có 2 nghiệm \(x=2;x=3\)
