1. It is known that $a+\frac{b^{2}}{a}=b+\frac{a^{2}}{b}$. Is it true that $a=b$?
1. Bring the left side to a common denominator: $a+$ $+\frac{b^{2}}{a}=\frac{a^{2}+b^{2}}{a}$. Similarly, we handle the right side. We obtain the equality:

$$
\frac{a^{2}+b^{2}}{a}=\frac{a^{2}+b^{2}}{b}
$$

From the condition, it is clear that $a \neq 0, b \neq 0$. Therefore, $a^{2}+b^{2}>0$ (see comment). Thus, both sides of the equality can be divided by $a^{2}+b^{2}$. We get $\frac{1}{a}=\frac{1}{b}$, from which $a=b$.

Comment. The following obvious statements are often used in solving problems (and in mathematics in general): 1) the square of a non-zero number is positive, 2) the sum of squares of several numbers is non-negative. If this sum is zero, then each of the numbers is zero.