radical in denomintor

If the entire denominator of a fraction is under a square root, then you can just multiply the numerator and denominator by that root. So in your case, you'd do this:

\(\\
\left( \frac{3}{\sqrt{x - 1}} \right) \left( \frac{ \sqrt{x-1} }{ \sqrt{x-1} } \right) = \frac{3 \sqrt{x-1} }{x - 1}\\
\)

That might not look any simpler than what you started with, but it's required in math to get roots out of denominators, so you need to do it.

It's only when you're adding a root to something else (or adding two roots, or subtracting) that you need to do something similar to what you suggested. So if you had this fraction:

\(\\
\frac{1}{1 + \sqrt{x}}\\
\)

then, because we're adding a root to something else, we'd need to use the difference of squares pattern (the "conjugate" if you've seen that word) to get the root out of the denominator. So here we'd multiply by 1 - √x in both the numerator and denominator:

\(\\
\frac{1}{1 + \sqrt{x}} = \left( \frac{1}{1 + \sqrt{x}} \right) \left( \frac{1 - \sqrt{x}}{1 - \sqrt{x}} \right) = \frac{1 - \sqrt{x}}{1 - x}\\
\)