\(\left(x-2\right)^4=\left(x-2\right)^8\)
TH1: \(x-2\ne0\Rightarrow x\ne2\)TA CÓ:
\(\left(x-2\right)^8:\left(x-2\right)^4=1\)
\(\Rightarrow\left(x-2\right)^4=1\)
\(\Rightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
TH2: \(x-2=0\Rightarrow x=2\)Ta có:
\(\left(2-2\right)^4=\left(2-2\right)^8\Rightarrow0=0\)(luôn đúng)
Vậy với \(x\in\left\{1;2;3\right\}\)thì \(\left(x-2\right)^4=\left(x-2\right)^8\)
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