## Task B-3.2.

Solve the equation: $\frac{x^{\frac{1}{4}}}{x^{\frac{1}{4}}+5}-\frac{5 x^{\frac{1}{4}}-15}{x^{\frac{1}{4}}-5}=\frac{50}{x^{\frac{1}{2}}-25}$.
## Solution.

Notice that the least common denominator of all fractions in the equation is

$$
\left(x^{\frac{1}{4}}+5\right)\left(x^{\frac{1}{4}}-5\right)=\left(x^{\frac{1}{4}}\right)^{2}-5^{2}=x^{\frac{1}{2}}-25 \text {. }
$$

and multiply the given equation by it. We get:

$$
x^{\frac{1}{4}}\left(x^{\frac{1}{4}}-5\right)-\left(5 x^{\frac{1}{4}}-15\right)\left(x^{\frac{1}{4}}+5\right)=50
$$

with the condition that $x \neq 625$.

Simplify the obtained expression. It follows step by step:

$$
\begin{gathered}
x^{\frac{1}{2}}-5 x^{\frac{1}{4}}-5 x^{\frac{1}{2}}-25 x^{\frac{1}{4}}+15 x^{\frac{1}{4}}+75=50 \\
-4 x^{\frac{1}{2}}-15 x^{\frac{1}{4}}+25=0
\end{gathered}
$$

Notice that $\left(x^{\frac{1}{4}}\right)^{2}=x^{\frac{1}{2}}$, meaning the obtained equation is quadratic. Since $x^{\frac{1}{4}} \geqslant 0$ for all real numbers $x$, the solution $x^{\frac{1}{4}}=-5$ is not possible.

The solution $x^{\frac{1}{4}}=\frac{5}{4}$ is positive, so the final solution is $x=\left(\frac{5}{4}\right)^{4}=\frac{625}{256}$.