Ta có : \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}\cdot\left[1-\left(x-7\right)^{x+10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{x+10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
Vậy \(x\in\left\{7;8\right\}\)
Ta có: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Rightarrow\left(x-7\right)^{\left(x+1\right)}.\left[1-\left(x-7\right)^{x+10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{x+10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{\left(x+10\right)}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)
Vậy...
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