Question 80: Given positive real numbers $x, y$ satisfying $xy=1$, try to find the range of values for $f(x, y)=\frac{x+y}{[x][y]+[x]+[y]+1}$, where $[\mathrm{x}]$ denotes the greatest integer less than or equal to $\mathrm{x}$.

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When $x=1$, $y=1$, at this time $f(1,1)=\frac{1}{2}$.
Assume $\mathrm{x}>1>\frac{1}{\mathrm{x}}=\mathrm{y}$. Let $[\mathrm{x}]=\mathrm{n}, \mathrm{x}=\mathrm{n}+\alpha(\alpha \in[0,1)$, then
$$
f(x, y)=\frac{(n+\alpha)+\frac{1}{n+\alpha}}{n+1}=1+\frac{\frac{1}{n+\alpha}(1-\alpha)}{n+1}=1+\frac{\frac{1}{n+\alpha}(1-\alpha)}{n+1}
$$

On one hand, $\frac{\frac{1}{n+\alpha}(1-\alpha)}{n+1} \leq \frac{\frac{1}{1+\alpha}(1-\alpha)}{1+1}=\frac{a^{2}}{2(1+\alpha)}=\frac{1}{2\left(\frac{1}{a^{2}+\frac{1}{a}}\right)}<\frac{1}{2(1+1)}=\frac{1}{4}$, so $f(x, y)<1+\frac{1}{4}=\frac{5}{4}$. When $x \rightarrow 2^{-}$, $f(x, y) \rightarrow \frac{5}{4}$.

On the other hand, $f(x, y)=\frac{(n+\alpha)+\frac{1}{n+\alpha}}{n+1} \geq \frac{n+\frac{1}{n}}{n+1}=\frac{n^{2}+1}{n(n+1)}=1-\frac{1}{3+(n-1)+\frac{2}{n-1}} \geq 1-\frac{1}{3+2+1}=\frac{5}{6}$, when $x=2$, the equality holds.
In summary, the range of $\mathrm{f}(\mathrm{x}, \mathrm{y})$ is $\left\{\frac{1}{2}\right\} \cup\left[\frac{5}{6}, \frac{5}{4}\right)$.