81. The difference $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$ is an integer. Find this number.
Solution. Since $40 \sqrt{2}-57<0$, then $|40 \sqrt{2}-57|=$ $=57-40 \sqrt{2}$. Then

$$
\begin{aligned}
A & =\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}= \\
& =\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{2}}
\end{aligned}
$$

Let $57-40 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a$ and $b$ are unknown coefficients. Then

$$
57-40 \sqrt{2}=a^{2}+2 b^{2}+2 \sqrt{2} a b
$$

from which

$$
\left\{\begin{array}{l}
a^{2}+2 b^{2}=57 \\
2 a b=-40
\end{array}\right.
$$

Solving this system of equations, we get $a=5, b=-4$. Therefore, $\sqrt{57-40 \sqrt{2}}=\sqrt{(5-4 \sqrt{2})^{2}}=|5-4 \sqrt{2}|=4 \sqrt{2}-5, \quad$ since $5-4 \sqrt{2}<0$.

Similarly, it is established that $\sqrt{57+40 \sqrt{2}}=4 \sqrt{2}+5$. Consequently, $A=4 \sqrt{2}-5-(4 \sqrt{2}+5)=-10$.

## Elimination of Irrationality in the Denominator