Task B-4.2. For what values of the real number $x$ in the expansion of $\left(\sqrt{5^{x}}+\frac{1}{\sqrt[3]{25^{x}}}\right)^{6}$ is the third term equal to 75?
Solution. The third term in the expansion of $\left(\sqrt{5^{x}}+\frac{1}{\sqrt[3]{25^{x}}}\right)^{6}$ is

$$
\begin{aligned}
& \binom{6}{2}\left(\sqrt{5^{x}}\right)^{4}\left(\frac{1}{\sqrt[3]{25^{x}}}\right)^{2} \\
& =15 \cdot 5^{2 x} \cdot 5^{-\frac{4 x}{3}}=3 \cdot 5^{1-\frac{2 x}{3}}
\end{aligned}
$$

Then $3 \cdot 5^{1-\frac{2 x}{3}}=75$, or $5^{1-\frac{2 x}{3}}=25$

so $1-\frac{2 x}{3}=2$ and finally $x=-\frac{3}{2}$.