Tìm x biết: (x-1)^x+2=(x-1)^x+4

Đáp án:

$x \in \left\{ {0;1;2} \right\}$

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