## Task B-2.1.

If $a=\sqrt{\sqrt{3}-\sqrt{2}}(\sqrt{\sqrt{3}-1}+\sqrt{\sqrt{3}+1}), b=\sqrt{\sqrt{3}+\sqrt{2}}(\sqrt{\sqrt{3}-1}-\sqrt{\sqrt{3}+1})$, what is $a+b$?
## First Solution.

$$
\begin{aligned}
a^{2} & =(\sqrt{3}-\sqrt{2})(\sqrt{3}-1+2 \sqrt{\sqrt{3}-1} \cdot \sqrt{\sqrt{3}+1}+\sqrt{3}+1) \\
& =2(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=2 \\
b^{2} & =(\sqrt{3}+\sqrt{2})(\sqrt{3}-1-2 \sqrt{\sqrt{3}-1} \cdot \sqrt{\sqrt{3}+1}+\sqrt{3}+1) \\
& =2(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=2
\end{aligned}
$$

Therefore, $|a|=\sqrt{2},|b|=\sqrt{2}$.

It is clear that $a>0$, and since $\sqrt{\sqrt{3}+1}>\sqrt{\sqrt{3}-1}$, it follows that $b<0$. Thus, from $a>0, b<0$ we have $a=\sqrt{2}, b=-\sqrt{2}$.

Finally, the sum $a+b=0$.

Note: If a student does not conclude that $b<0$ and obtains the solution $a+b=2 \sqrt{2}$, deduct 2 points.