\(\left(x-5\right)^4+\left(x-3\right)^4=16\)
Đặt \(x-4=t\)
\(\Leftrightarrow\left(t-1\right)^4+\left(t+1\right)^4=16\\ \Leftrightarrow\left(t-1\right)\left(t^3-3t^2+3t-1\right)+t^4+4t^3+6t^2+4t+1=16\\ \Leftrightarrow t^4-3t^3+3t^2-t-t^3+3t^2-3t+1+t^4+4t^3+6t^2+4t+1-16=0\\ \Leftrightarrow2t^4+12t^2-14=0\\ \Leftrightarrow t^4+6t^2-7=0\\ \Leftrightarrow t^4+7t^2-t^2-7=0\\ \Leftrightarrow\left(t^4+7t^2\right)-\left(t^2+7\right)=0\\ \Leftrightarrow t^2\left(t^2+7\right)-\left(t^2+7\right)=0\\ \Leftrightarrow\left(t^2+7\right)\left(t^2-1\right)=0\\ \Leftrightarrow\left(t+1\right)\left(t-1\right)=0\left(\text{Vì }t^2+7\ne0\right)\\ \Leftrightarrow\left(x-4+1\right)\left(x-4-1\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\\ \)
Vậy tập nghiệm phương trình là \(S=\left\{3;5\right\}\)
