`1) ( x - 1 )^{x + 2} = ( x - 1 )^2`
⇔ \(\left[ \begin{array}{l}x-1=0\\x+2=2\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=0+1\\x=2-2\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=1\\x=0\end{array} \right.\)
Vậy `x ∈ { 0 ; 1 }`
`2) ( x + 3 )^{y + 1} = ( 2x - 1 )^{y-1}`
`⇔ x + 3 = 2x - 1`
`⇔ 2x - x = 3 + 1`
`⇔ x = 4`
Vậy `x = 4`