Solve the following equation:

$$
\sqrt{\frac{x}{a}+\frac{c}{b}}+\sqrt{\frac{x}{a}-\frac{c}{b}}=\sqrt{\frac{4 x}{a}-\frac{2 c}{b}}
$$
Let's consider the following expressions:

$$
\begin{aligned}
f_{1} & =\sqrt{\frac{x}{a}+\frac{c}{b}}+\sqrt{\frac{x}{a}+\frac{c}{b}}+\sqrt{\frac{4 x}{a}-\frac{2 c}{b}} \\
f_{2} & =\sqrt{\frac{x}{a}+\frac{c}{b}}+\sqrt{\frac{x}{a}-\frac{c}{b}}+\sqrt{\frac{4 x}{a}-\frac{2 c}{b}}
\end{aligned}
$$

Since we are considering real-valued square roots, \( f_{1} \) and \( f_{2} \) are real at the same time. However, \( f_{1} \) is positive, while \( f_{2} \) can be zero, and the question is for which value of \( x \) will \( f_{2} = 0 \)? If \( f_{2} = 0 \), then \( f_{1} f_{2} = 0 \) and conversely: if \( f_{1} f_{2} = 0 \), then \( f_{2} = 0 \) (since \( f_{1} > 0 \)). Therefore, the equation \( f_{2} = 0 \) is equivalent to the equation

$$
f_{1} f_{2}=\frac{x}{a}+\frac{c}{b}+\frac{x}{a}-\frac{c}{b}+2 \sqrt{\frac{x^{2}}{a^{2}}-\frac{c^{2}}{b^{2}}}-\left(\frac{4 x}{a}-\frac{2 c}{b}\right)=0
$$

or, after rearranging:

$$
f_{1} f_{2}=-\frac{2 x}{a}+\frac{2 c}{b}+2 \sqrt{\frac{x^{2}}{a^{2}}-\frac{c^{2}}{b^{2}}}=0
$$

or

$$
f_{1} f_{2}=2\left[\sqrt{\left(\frac{x}{a}+\frac{c}{b}\right)\left(\frac{x}{a}-\frac{c}{b}\right)}-\left(\frac{x}{a}-\frac{c}{b}\right)\right]=0
$$

If \(\frac{x}{a}-\frac{c}{b} > 0\), then

$$
f_{1} f_{2}=2 \sqrt{\frac{x}{a}-\frac{c}{b}}\left(\sqrt{\frac{x}{a}+\frac{c}{b}}-\sqrt{\frac{x}{a}-\frac{c}{b}}\right)=0
$$

1. \( f_{1} f_{2} = 0 \), if \(\frac{x}{a}-\frac{c}{b} = 0\); in this case, \( f_{2} = 0 \), but \( f_{1} \neq 0 \) and \( x = \frac{a c}{b} \).

That is,

$$
\text{if } \frac{x}{a} = \frac{c}{b}, \text{ then } f_{2} = \sqrt{\frac{2 c}{b}} - \sqrt{\frac{2 c}{b}} = 0
$$

while

$$
f_{1} = \sqrt{\frac{2 c}{b}} + \sqrt{\frac{2 c}{b}} = 2 \sqrt{\frac{2 c}{b}}
$$

2. \( f_{1} f_{2} = 0 \), if

$$
\sqrt{\frac{x}{a}+\frac{c}{b}} = \sqrt{\frac{x}{a}-\frac{c}{b}}
$$

This equality becomes an identity if \(\frac{c}{b} = 0\), and thus \( f_{2} = 0 \) for any value of \( x \), while \( f_{1} = 4 \sqrt{\frac{x}{a}} \). However, if \(\frac{c}{b} \neq 0\), it contains a contradiction and does not provide a finite solution.