\(\left(5-x\right)^6\)=\(\left(x-2\right)^6\)
ko tồn tại x
\(\left(5-x\right)^6=\left(x-2\right)^6\)
\(\Leftrightarrow\orbr{\begin{cases}5-x=2-x\\x-5=2-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0=-3\left(L\right)\\2x=7\end{cases}}\Leftrightarrow x=\frac{7}{2}\)
\(\left(5-x\right)^6=\left(x-2\right)^6\)
\(\Rightarrow\orbr{\begin{cases}5-x=x-2\\5-x=2-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5+2=2x\\x\in\varnothing\end{cases}}\)
\(\Rightarrow x=\frac{7}{2}\)