$11.15 \sqrt{\frac{1+x}{x}}+\frac{1}{x}=5$
11.15 Here $\frac{x+1}{x} \geq 0$. Solving this inequality using the interval method, we find $x \leq -1$ or $x > 0$. Let's rewrite the given equation as: $\sqrt{\frac{1}{x}+1}+\frac{1}{x}=5$. By setting $\frac{1}{x}=y$, we get $\sqrt{y+1}+y=5$, or

$\sqrt{y+1}=5-y$.

The domain of the equation (1) is $y \geq -1$: in this domain $\sqrt{y+1} \geq 0$; therefore, $5-y \geq 0$, i.e., $y \leq 5$. Thus, the sought root belongs to the interval $-1 \leq y \leq 5$. Squaring both sides of equation (1):

\[
\begin{aligned}
& y+1=25-10 y+y^{2} ; y^{2}-11 y+24=0 \\
& y_{1}=3 \in[-1,5] ; y_{2}=8 \notin[-1,5]
\end{aligned}
\]

Finally, solving the equation $\frac{1}{x}=3$, we get $x=\frac{1}{3}>0-$ the root of the given equation.

Answer: $x=\frac{1}{3}$.

Remarks. 1. It is useful to check the obtained root. 2. All the indicated actions can be performed without finding the domain of the equation, but it is important to remember that squaring both sides of the equation expands its domain and extraneous roots may appear. After such a solution, it is mandatory to check the roots by substituting them into the original equation.