## Task B-2.1.

Calculate $\sqrt{1-A^{-1}}$ if $A=\frac{\sqrt{2022}+1}{\sqrt{2022}-1}-\frac{2}{\sqrt{2022}+1}$.
## Solution.

By reducing to a common denominator, we will calculate the value of the expression $A$.

$$
\begin{array}{cc}
A=\frac{\sqrt{2022}+1}{\sqrt{2022}-1}-\frac{2}{\sqrt{2022}+1}=\frac{(\sqrt{2022}+1)^{2}-2(\sqrt{2022}-1)}{(\sqrt{2022}-1)(\sqrt{2022}+1)} & 1 \text{point} \\
=\frac{2022+2 \sqrt{2022}+1-2 \sqrt{2022}+2}{2022-1} & 2 \text{ points} \\
=\frac{2025}{2021} . & 1 \text{point}
\end{array}
$$

Then we have:

$$
\sqrt{1-A^{-1}}=\sqrt{1-\frac{2021}{2025}}=\sqrt{\frac{4}{2025}}=\frac{2}{45}
$$