$2.331 A=\frac{\sqrt{3} x^{3 / 2}-5 x^{1 / 3}+5 x^{4 / 3}-\sqrt{3 x}}{\sqrt{3 x+10 \sqrt{3} x^{5 / 6}+25 x^{2 / 3}} \cdot \sqrt{1-2 x^{-1}+x^{-2}}}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.

$2.331 A=\frac{\sqrt{3} x^{3 / 2}-5 x^{1 / 3}+5 x^{4 / 3}-\sqrt{3 x}}{\sqrt{3 x+10 \sqrt{3} x^{5 / 6}+25 x^{2 / 3}} \cdot \sqrt{1-2 x^{-1}+x^{-2}}}$.

The equation remains the same in English as it is a mathematical expression.
Solution. Obviously, $x>0$. Consider the numerator of the fraction $A$.

$B=\sqrt{3} x^{3 / 2}-\sqrt{3} x^{1 / 2}+5 x^{4 / 3}-5 x^{1 / 3}=\sqrt{3} x^{1 / 2}(x-1)+5 x^{1 / 3}(x-1)=(x-1)(\sqrt{3 x}+5 \sqrt[3]{x})$.

Consider the factors of the denominator of $A$ one by one.

$C=\sqrt{3 x+2 \cdot 5 \cdot \sqrt{3} x^{1 / 2} \cdot x^{1 / 3}+25 x^{2 / 3}}=\sqrt{(\sqrt{3 x}+5 \sqrt[3]{x})^{2}}=$ $=|\sqrt{3 x}+5 \sqrt[3]{x}|=\sqrt{3 x}+5 \sqrt[3]{x}$, since $x>0$.

$D=\sqrt{1-\frac{2}{x}+\frac{1}{x^{2}}}=\sqrt{\left(1-\frac{1}{x}\right)^{2}}=\left|1-\frac{1}{x}\right|=\frac{|x-1|}{|x|}=\frac{|x-1|}{x}$.

We get $A=\frac{B}{C \cdot D}=\frac{x-1}{|x-1|} \cdot|x|$.

1) $\left\{\begin{array}{l}0<x<1, \\ A=-x .\end{array}\right.$

2) $\left\{\begin{array}{l}x>1, \\ A=x .\end{array}\right.$

Answer: if $0<x<1$, then $A=-x$; if $x>1$, then $A=x$.