\[\begin{array}{l}
A = \frac{{5\sqrt x }}{{x - 1}} + \frac{3}{{2\sqrt x + 2}} - \frac{5}{{2\sqrt x - 2}}\\
dk:x \ge 0;x \ne 1\\
= \frac{{5\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \frac{3}{{2\left( {\sqrt x + 1} \right)}} - \frac{5}{{2\left( {\sqrt x - 1} \right)}}\\
= \frac{{10\sqrt x + 3\left( {\sqrt x - 1} \right) - 5\left( {\sqrt x + 1} \right)}}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{10\sqrt x + 3\sqrt x - 3 - 5\sqrt x - 5}}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{8\sqrt x - 8}}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{8\left( {\sqrt x - 1} \right)}}{{2\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{4}{{\sqrt x + 1}}
\end{array}\]

Vậy với \[x \ge 0;x \ne 1\] thì \[A = \frac{4}{{\sqrt x + 1}}\]