Đáp án:
$x \in \left\{ {0;1;2} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
{\left( {x - 1} \right)^{x + 2}} = {\left( {x - 1} \right)^{x + 4}}\\
\Leftrightarrow {\left( {x - 1} \right)^{x + 2}} = {\left( {x - 1} \right)^{x + 2}}.{\left( {x - 1} \right)^2}\\
\Leftrightarrow {\left( {x - 1} \right)^{x + 2}}\left[ {{{\left( {x - 1} \right)}^2} - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {x - 1} \right)^{x + 2}} = 0\\
{\left( {x - 1} \right)^2} - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
{\left( {x - 1} \right)^2} = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0
\end{array} \right.\\
vay\,x \in \left\{ {0;1;2} \right\}
\end{array}$
