Solve the equation

$$
\frac{x^{2}+1}{x^{2}+11}=\frac{1}{6} \sqrt{\frac{11 x-6}{6-x}}
$$
Solution. Due to the interpretability of the fraction and square root expression in the equation, $x \neq 6$ and $\frac{11 x-6}{6-x} \geq 0$, which conditions are satisfied when $x \in\left[\frac{6}{11} ; 6[\right.$.

Multiplying the equation by 6:

$$
\frac{6\left(x^{2}+1\right)}{x^{2}+11}=\sqrt{\frac{11 x-6}{6-x}}
$$

The function

$$
f(x)=\frac{6\left(x^{2}+1\right)}{x^{2}+11}=6\left(1-\frac{10}{x^{2}+11}\right)
$$

is strictly monotonically increasing in its domain, hence it is invertible, and its inverse is

$$
f^{-1}(x)=\sqrt{\frac{11 x-6}{6-x}}
$$

Thus, the expression on the right side of the equation is the inverse of the left side.

Since the image of a function and its inverse is symmetric with respect to the line $y=x$, equality can only hold when $y=x$. Therefore, our original equation is equivalent to

$$
\frac{6\left(x^{2}+1\right)}{x^{2}+11}=x
$$

Rearranging: $0=x^{3}-6 x^{2}+11 x-6$. Factoring:

$$
0=(x-1)(x-2)(x-3)
$$

Thus, $x_{1}=1, x_{2}=2, x_{3}=3$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1df94dc7b1de503b2779g-1.jpg?height=450&width=585&top_left_y=1222&top_left_x=759)

The obtained roots satisfy the equation, as they all belong to the domain of $f$.