2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.

2.030. $\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}$.

(Note: The original text and the translation are identical as the expression is already in a mathematical form that does not require translation.)
Solution.

Domain of definition: $\left\{\begin{array}{l}a>0, \\ a \neq \sqrt{2} .\end{array}\right.$

$$
\sqrt{\frac{\sqrt{2}}{a}+\frac{a}{\sqrt{2}}+2}-\frac{a^{2} \sqrt[4]{2}-2 \sqrt{a}}{a \sqrt{2 a}-\sqrt[4]{8 a^{4}}}=\sqrt{\frac{a^{2}+2 a \sqrt{2}+(\sqrt{2})^{2}}{a \sqrt{2}}}-
$$

$$
\begin{aligned}
& -\frac{a^{2} \cdot 2^{1 / 4}-2 \cdot a^{1 / 2}}{a^{3 / 2} \cdot 2^{1 / 2}-2^{3 / 4} \cdot a}=\frac{\sqrt{(a+\sqrt{2})^{2}}}{\sqrt{a \sqrt{2}}}-\frac{2^{1 / 4} \cdot a^{1 / 2} \cdot\left(a^{3 / 2}-2^{3 / 4}\right)}{2^{1 / 2} \cdot a \cdot\left(a^{1 / 2}-2^{1 / 4}\right)}= \\
& =\frac{a+\sqrt{2}}{a^{1 / 2} \cdot 2^{1 / 4}}-\frac{2^{1 / 4} \cdot a^{1 / 2} \cdot\left(\left(a^{1 / 2}\right)^{3}-\left(2^{1 / 4}\right)^{3}\right)}{2^{1 / 2} \cdot a \cdot\left(a^{1 / 2}-2^{1 / 4}\right)}=\frac{a+2^{1 / 2}}{a^{1 / 2} \cdot 2^{1 / 4}}- \\
& -\frac{\left(a^{1 / 2}-2^{1 / 4}\right)\left(a+2^{1 / 4} \cdot a^{1 / 2}+2^{1 / 2}\right)}{a^{1 / 2} \cdot 2^{1 / 4} \cdot\left(a^{1 / 2}-2^{1 / 4}\right)}=\frac{a+2^{1 / 2}}{a^{1 / 2} \cdot 2^{1 / 4}}-\frac{a+2^{1 / 4} \cdot a^{1 / 2}+2^{1 / 2}}{a^{1 / 2} \cdot 2^{1 / 4}}= \\
& =\frac{a+2^{1 / 2}-a-2^{1 / 4} \cdot a^{1 / 2}-2^{1 / 2}}{a^{1 / 2} \cdot 2^{1 / 4}}=\frac{-2^{1 / 4} \cdot a^{1 / 2}}{a^{1 / 2} \cdot 2^{1 / 4}}=-1
\end{aligned}
$$

Answer: -1.