\(\left(2x-4\right)^{38}=\left(2x-4\right)^{48}\)
\(\Rightarrow\left(2x-4\right)^{38}-\left(2x-4\right)^{48}=0\)
\(\Rightarrow\left(2x-4\right)^{38}\left[1-\left(2x-4\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-4\right)^{38}=0\\1-\left(2x-4\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x-4=1\\2x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}.\)
#\(Toru\)
