1. Solve in the set of real numbers the equation: $(x-1)^{x+2}=(x-1)^{x^{2}+x}$.
1. Solve in the set of real numbers the equation: $(x-1)^{x+2}=(x-1)^{x^{2}+x}$. Solution

If $x-1=-1$, we have $x=0$ and $(-1)^{2}=(-1)^{0}$ true. $\qquad$
If $x-1=0$, we have $x=1$ and $0^{3}=0^{2} \quad$ true.

![](https://cdn.mathpix.com/cropped/2024_06_07_6471d89924ab72f11e5ag-2.jpg?height=179&width=1405&top_left_y=410&top_left_x=1512)$\qquad$
If $x-1=1$, we have $x=2$ and $1^{4}=1^{6} \quad$ true.

$1 p$

If $x-1$ Ï $\{-1,0,1\}$, then $x+2=x^{2}+x$ P $\quad x= \pm \sqrt{2}$ $\qquad$
If $x=-\sqrt{2}$, then $x-10$, so $(x-1)^{x+2}$ exists $\qquad$
$1 p$

Thus the set of solutions is $\{0,1,2, \sqrt{2}\}$. $\qquad$
If $z=a+b i, a, b \hat{I}$. continuation