1. Find all values of $x$, for each of which one of the three given numbers $\log _{x^{2}}\left(x^{2}-3 x+2\right)$, $\log _{x^{2}} \frac{x^{2}}{x-2}$, and $\log _{x^{2}} \frac{x^{2}}{x-1}$ is equal to the sum of the other two.
Answer: $x=3$.

Solution. Note that on the domain of definition, the sum of all three logarithms is

$$
\log _{x^{2}}\left(\frac{x^{2}}{x-2} \cdot \frac{x^{2}}{x-1}\left(x^{2}-3 x+2\right)\right)=\log _{x^{2}} x^{4}=2
$$

Let the number that is equal to the sum of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a+b$ and $a+b+c=2$, from which it follows that $c=1$, i.e., one of the three given logarithms is equal to 1.

The converse is also true, namely, if one of the three given logarithms is equal to 1, then since the sum of all three logarithms is 2, the sum of the remaining two is 1, i.e., their sum is equal to the first logarithm.

Thus, the condition of the problem is satisfied if and only if one of the logarithms is equal to 1 (and all logarithms exist). A logarithm is equal to 1 when its base is equal to the logarithmic expression. We obtain the system

$$
\left[\begin{array}{l}
x^{2}=\frac{x^{2}}{x-2}, \\
x^{2}=\frac{x^{2}}{x-1}, \\
x^{2}=x^{2}-3 x+2
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=0 \\
x=3 \\
x=2 \\
x=\frac{2}{3}
\end{array}\right.\right.
$$

Of the found values of the variable, only $x=3$ satisfies the domain of definition.