Let's find the binomial such that in its 7th power, the third term is $6 \frac{2}{9} a^{5}$, and the fifth term is $52 \frac{1}{2} a^{3}$.
The questionable binomial will be of the form $(x+a y)$.

In the context of the problem,

$$
\binom{7}{2} x^{2} a^{5} y^{5}=6 \frac{2}{9} a^{5}
$$

$$
\binom{7}{4} x^{4} a^{3} y^{3}=52 \frac{1}{2} a^{3}
$$

From (1),

$$
x^{2}=\frac{8}{27 y^{5}}
$$

and from (2),

$$
x^{4}=\frac{3}{2 y^{3}}
$$

thus,

$$
\left(\frac{8}{27 y^{5}}\right)^{2}=\frac{3}{2 y^{3}}
$$

or

$$
y^{7}=\frac{2^{7}}{3^{7}}
$$

so the real value of $y$ is $\frac{2}{3}$. Substituting this value into (1), we get $x=\frac{3}{2}$.

Therefore, $\left(\frac{3}{2}+\frac{2}{3} a\right)$ is the sought binomial.

(Endre Sárközy, Keszthely.)

The problem was also solved by: Bartók I., Deutsch E., Deutsch I., Enyedi B., Haar A., Harsányi Z., Heimlich P., Hirschfeld Gy., Jánosy Gy., Kertész G., Kiss J., Kürti I., Liebner A., Ligeti P., Neidenbach E., Pám M., Pichler S., Pivnyik I., Popoviciu M., Preisich G., Rássy P., Riesz K., Riesz M., Schlesinger O., Schuster Gy., Schwemmer I., Szántó H., Szücs A., Tóth B., Weisz P.