ĐK:\(x\in Q;xe0\)
Ta có: \(1^x=1\forall x\)
\(\Rightarrow x⋮x^{10}=1\)
\(\Rightarrow x=x^{10}\)
\(\Rightarrow x^{10}-x=0\)
\(\Rightarrow x.\left(x^9-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^9-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\x^9=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) mà \(xe0\)
Vậy \(x\) =1
