Solve the following equation: $x^{2}+\left(\frac{5 x}{x-5}\right)^{2}=11$.
I. solution. We bring to a common denominator and transform the numerator:

$$
\frac{x^{4}-10 x^{2}(x-5)}{(x-5)^{2}}=11
$$

This equation can also be written in the following form:

$$
\left(\frac{x^{2}}{x-5}\right)^{2}-10 \cdot \frac{x^{2}}{x-5}-11=0
$$

We apply the substitution $y=\frac{x^{2}}{x-5}$: $y^{2}-10 y-11=0$, whose roots are: $y_{1}=11$ and $y_{2}=-1$.

1. case: $\frac{x^{2}}{x-5}=11$, from which $x^{2}-11 x+55=0$; this quadratic equation has no real roots, as its discriminant is negative.
2. case: $\frac{x^{2}}{x-5}=-1$, from which $x^{2}+x-5=0$. Then $x_{1,2}=\frac{-1 \pm \sqrt{21}}{2}$.

These numbers are also roots of the original equation, because our transformations are equivalent.

II. solution. The domain of definition (due to the denominator): $x \neq 5$. After multiplying by $(x-5)^{2}$ and rearranging, we get:

$$
x^{4}-10 x^{3}+39 x^{2}+110 x-275=0
$$

The expression on the left side of the equation can be factored:

$$
\left(x^{2}+x-5\right)\left(x^{2}-11 x+55\right)=0
$$

On the left side, there is a product of two factors. A product is zero if and only if one of its factors is zero, so we examine two cases.

1. case: If $x^{2}+x-5=0$, then $x_{1,2}=\frac{-1 \pm \sqrt{21}}{2}$.
2. case: If $x^{2}-11 x+55=0$, then the discriminant is: $D=121-220<0$. This equation has no real solutions.

Therefore, the solutions of the original equation are: $x_{1}=\frac{-1+\sqrt{21}}{2}, x_{2}=\frac{-1-\sqrt{21}}{2}$.