Could someone elaborate on this algebraic transformation?

Question

I have a solution for an exercise and one part of it is not clear to me:

$ \frac{\frac{1}{\sqrt{x - 1}} - 1}{x - 2} = \frac{1 - \sqrt{x - 1}}{(x - 2)\sqrt{x - 1}} $

Could anyone explain, please, how the result was obtained?

The full solution is this:

$ \frac{\frac{1}{\sqrt{x - 1}} - 1}{x - 2} = \frac{1 - \sqrt{x - 1}}{(x - 2)\sqrt{x - 1}} * \frac{1 + \sqrt{x - 1}}{1 + \sqrt{x - 1}} = \frac{2 - x}{(x - 2)\sqrt{x - 1}(1 + \sqrt{x - 1})} = \frac{-1}{\sqrt{x - 1}(1 + \sqrt{x - 1})}, x \ne 2 $

Answer 1

You simply multiply the top and bottom by $\sqrt{x-1}$:

$$\frac{\frac{1}{\sqrt{x-1}}-1}{x-2}=\frac{\Bigl(\frac{1}{\sqrt{x-1}}-1\Bigr)\sqrt{x-1}}{(x-2)\sqrt{x-1}}=\frac{1-\sqrt{x-1}}{(x-2)\sqrt{x-1}}$$

Edit: This is valid as long as the thing you are multiplying top and bottom by is not $0$. In this case, $\sqrt{x-1} \neq 0$ for or else the thing you were given containing a $\frac{1}{\sqrt{x-1}}$ would have been undefined in the first place.

Answer 2

In my retirement, I have been going to local high-schools as a volunteer, and I find, first, that the students have never been told this technique of simplifying a complicated fraction, and, second (much more discouragingly), that the teachers not only are unfamiliar with it, but also are unwilling to apply it.

Illustrative example: $$ \frac{\frac12+\frac34}{\frac23-\frac16}=\frac{6+9}{8-2}=\frac{15}6=\frac52\,, $$ the first step being achieved by multiplying top and bottom by $12$.