$7.10 \quad 10^{\frac{2}{x}}+25^{\frac{1}{x}}=4.25 \cdot 50^{\frac{1}{x}}$.
7.10 Let's preliminarily perform the transformations:

$$
\begin{aligned}
10^{\frac{2}{x}} & =(2 \cdot 5)^{\frac{2}{x}}=2^{\frac{2}{x}} \cdot 5^{\frac{2}{x}} ; 25^{\frac{1}{x}}=\left(5^{2}\right)^{\frac{1}{x}}=5^{\frac{2}{x}} \\
50^{\frac{1}{x}} & =\left(2 \cdot 5^{2}\right)^{\frac{1}{x}}=2^{\frac{1}{x}} \cdot 5^{\frac{2}{x}}
\end{aligned}
$$

Dividing both sides of the given equation by $5^{\frac{2}{x}} \neq 0$, we obtain the equation

$2^{\frac{2}{x}}+1=4.25 \cdot 2^{\frac{1}{x}}$,

which we solve by substitution $y=2^{\frac{1}{x}}$.

Answer: $x_{1,2}= \pm 0.5$.