$2.326 A=\frac{\sqrt{a+2 \sqrt{b}+\frac{b}{a}} \cdot \sqrt{2 a-10 \sqrt[6]{8 a^{3} b^{2}}+25 \sqrt[3]{b^{2}}}}{a \sqrt{2 a}+\sqrt{2 a b}-5 a \sqrt[3]{b}-5 \sqrt[6]{b^{5}}}$.

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$2.326 A=\frac{\sqrt{a+2 \sqrt{b}+\frac{b}{a}} \cdot \sqrt{2 a-10 \sqrt[6]{8 a^{3} b^{2}}+25 \sqrt[3]{b^{2}}}}{a \sqrt{2 a}+\sqrt{2 a b}-5 a \sqrt[3]{b}-5 \sqrt[6]{b^{5}}}$.

The above text is already in English and does not require translation.
Solution. Note that $a>0, b \geq 0$. Therefore,

$$
\begin{aligned}
& \sqrt{a+2 \sqrt{b}+\frac{b}{a}}=\sqrt{\frac{a^{2}+2 a \sqrt{b}+b}{a}}=\sqrt{\frac{(a+\sqrt{b})^{2}}{a}}=\frac{|a+\sqrt{b}|}{\sqrt{a}}=\frac{a+\sqrt{b}}{\sqrt{a}} \\
& \sqrt{2 a-10 \sqrt[6]{8 a^{3} b^{2}}+25 \sqrt[3]{b^{2}}}=\sqrt{2 a-10 \cdot \sqrt{2 a} \cdot \sqrt[3]{b}+25 \sqrt[3]{b^{2}}}= \\
& =\sqrt{(\sqrt{2 a}-5 \sqrt[3]{b})^{2}}=|\sqrt{2 a}-5 \sqrt[3]{b}|
\end{aligned}
$$

Transform the denominator of the fraction $A$.

$a \sqrt{2 a}+\sqrt{2 a b}-5 a \sqrt[3]{b}-5 \sqrt[6]{b^{5}}=\sqrt{2 a}(a+\sqrt{b})-5 \sqrt[3]{b}(a+\sqrt{b})=(a+\sqrt{b})(\sqrt{2 a}-5 \sqrt[3]{b})$. Then

$$
A=\frac{(a+\sqrt{b})}{\sqrt{a}} \cdot \frac{|\sqrt{2 a}-5 \sqrt[3]{b}|}{(a+\sqrt{b})(\sqrt{2 a}-5 \sqrt[3]{b})}=\frac{|\sqrt{2 a}-5 \sqrt[3]{b}|}{\sqrt{a}(\sqrt{2 a}-5 \sqrt[3]{b})}
$$

$\left\{\begin{array}{l}\sqrt{2 a}5 \sqrt[3]{b} \\ A=\frac{1}{\sqrt{a}}\end{array}\right.$

Answer: if $\sqrt{2 a}5 \sqrt[3]{b}$, then $A=\frac{1}{\sqrt{a}}$.