Let's find all non-negative real numbers $A, B$ such that for every $x > 1$, the following holds:

$$
\left[\frac{1}{A x+\frac{B}{x}}\right]=\frac{1}{A \cdot[x]+\frac{B}{[x]}}
$$
The expression on the right side cannot be zero for any $x$. However, the left side can be zero, specifically when $A x + B / x > 1$. If $A \neq 0$, there exists an $x_{0}$ such that $A x_{0} + B / x_{0} > 1$ and the right side is defined at $x_{0}$. Therefore, only $A=0$ is possible. Let's substitute $A=0$ into the equation:

$$
\left[\frac{x}{B}\right]=\frac{[x]}{B}
$$

If $B$ is not 1, the left side is an integer, but the right side is not always an integer. Therefore, the solution is $A=0, B=1$. In this case, both the left and right sides are $[x]$, so they are indeed equal for all $x$ greater than 1.