\(x^{28}=x\)
\(x^{27}.x-x=0\)
\(x^{27}.\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-1=0\\x^{27}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
Vậy x=1 hoặc x=0
\(x^{28}=x\)
\(x^{28}-x=0\)
\(x^{27}.x-x=0\)
\(x\left(x^{27}-1\right)=0\)
\(\orbr{\begin{cases}x=0\\x^{27}-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x^{27}=1\Rightarrow x=1\end{cases}}\)