Tìm x biết: (x – 1)^x+2 = (x – 1)^x+4

(x – 1)^x+2 = (x – 1)^x+4

⇔ (x – 1)^x+2 – (x – 1)^x+4 = 0

⇔ (x – 1)^x+2[1 – (x – 1)²] = 0

⇔ \(\left[ \begin{array}{l}{\left( {x - 1} \right)^{x + 2}} = 0\\1 - {\left( {x - 1} \right)^2} = 0\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x - 1 = 0\\{\left( {x - 1} \right)^2} = 1\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}x = 1\\x = 2\\x = 0\end{array} \right.\)

Vậy x ∈ {0; 1; 2}