[tex]a=\sqrt7-\sqrt5\\
b=\sqrt5-\sqrt3\\\\
a)A^{-1}\ \textgreater \ B^{-1}\\
\frac{1}{\sqrt7-\sqrt5}\ \textgreater \ \frac{1}{\sqrt5-\sqrt3}\\
\frac{\sqrt7+\sqrt5}{2}\ \textgreater \ \frac{\sqrt5+\sqrt3}{2}|\cdot 2\\
\sqrt7+\sqrt5\ \textgreater \ \sqrt5+\sqrt3\\
\sqrt7\ \textgreater \ \sqrt3 |()^2\\
7\ \textgreater \ 3 (A)\\
\\
b)\frac{1}{a}-\frac{1}{b}=\frac{a+b}{2}\\
\frac{1}{\sqrt7-\sqrt5}-\frac{1}{\sqrt5-\sqrt3}=\frac{\sqrt7-\sqrt5+\sqrt5-\sqrt3}{2}\\
\frac{\sqrt7+\sqrt5}{2}-\frac{\sqrt5+\sqrt3}{2}=\frac{\sqrt7-\sqrt3}{2}\\
\frac{\sqrt7+\sqrt5-\sqrt5-\sqrt3}{2}=\frac{\sqrt7-\sqrt3}{2}\\
[/tex]
[tex]\frac{\sqrt7-\sqrt3}{2}=\frac{\sqrt7-\sqrt3}{2}(A)[/tex]