2. Find all real numbers $x$ for which

$$
\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}=1
$$
IV/2. Let's recall that $a^{b}=1$ if $a=1$ or $a=-1$ and $b$ is an even number, or if $b=0$ and $a \neq 0$.

First, let's assume that the base $x^{2}-7 x+11$ is equal to 1. The equation $x^{2}-7 x+11=1$ can be rewritten as $x^{2}-7 x+10=0$, and the solutions are $x_{1}=2$ and $x_{2}=5$.

Next, we find the values of $x$ for which the base $x^{2}-7 x+11$ is equal to -1: the equation $x^{2}-7 x+11=$ -1 can be rewritten as $x^{2}-7 x+12=0$, which has solutions $x_{3}=3$ and $x_{4}=4$. The exponent $x^{2}+5 x-6$ is even for both of these values: for $x_{3}$ it is $9+15-6=18$, and for $x_{4}$ it is $16+20-6=30$. This means that for $x=3$ and $x=4$, $\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}=1$.

Finally, we find the values for which the exponent $x^{2}+5 x-6$ is equal to 0. The equation $x^{2}+$ $5 x-6=0$ has solutions $x_{5}=1$ and $x_{6}=-6$. Both solutions are valid, as the base is not equal to 0 in these cases. The expression $\left(x^{2}-7 x+11\right)^{x^{2}+5 x-6}$ has a value of 1 if $x$ is equal to $1, 2, 3, 4, 5$ or -6.

Solutions $x_{1}=2$ and $x_{2}=5$, when the base $x^{2}-7 x+11$ is equal to $1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots 2$ points

Solutions $x_{3}=3$ and $x_{4}=4$, when the base is equal to -1 and the exponent is an even number . . 2 points

Solutions $x_{5}=1$ and $x_{6}=-6$, when the exponent is equal to $0 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$.

(If the contestant correctly determines $x_{5}$ and $x_{6}$ but does not verify that the base is non-zero, deduct 1 point.)