$1.40 \frac{(x+1)^{-\frac{1}{2}}}{(x-1)^{-\frac{1}{2}}-(x+1)^{-\frac{1}{2}}} ; x=\frac{a^{2}+1}{2 a}$.

$1.41\left(\frac{\left(x+\sqrt[3]{2 a x^{2}}\right)\left(2 a+\sqrt[3]{4 a^{2} x}\right)^{-1}-1}{\sqrt[3]{x}-\sqrt[3]{2 a}}-(2 a)^{-\frac{1}{3}}\right)^{-6}$.
1.40 Here $x>1$, therefore

$$
\frac{a^{2}+1}{2 a}>1 \Rightarrow \frac{a^{2}-2 a+1}{2 a}>0 \Rightarrow \frac{(a-1)^{2}}{2 a}>0 \Rightarrow a>0
$$

Next, we find

$$
x+1=\frac{(a+1)^{2}}{2 a} ; x-1=\frac{(a-1)^{2}}{2 a}
$$

Then the given expression will take the following form:

$\frac{(a+1)^{-1} \sqrt{2 a}}{\frac{\sqrt{2 a}}{|a-1|}-\frac{\sqrt{2 a}}{a+1}}=\frac{|a-1|}{a+1-|a-1|}$.

## From this we obtain

for $a \in(0,1) \Rightarrow \frac{-(a-1)}{a+1+a-1}=\frac{1-a}{2 a}$;

for $a \in(1, \infty) \Rightarrow \frac{a-1}{2}$.

Answer: for $a \in(0,1) \Rightarrow \frac{1-a}{2 a} ;$ for $a \in(1, \infty) \Rightarrow \frac{a-1}{2}$.