# Rationalizing $\frac{\left(\frac{1}{\sqrt{x}}\right)-1}{x-1}$

I'm working through some rationalization problems and came across this problem:

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

$\frac{-1}{\sqrt{x}+x}$

I can't for the life of me figure out how to get there. Pretty sure I'm supposed to multiply $\frac{1}{\sqrt{x}}-1$ by its conjugate radical $\frac{1}{\sqrt{x}}+1$, but I still can't get to their answer.

A step-by-step rationalization would be much appreciated!

• Hint: $x-1=\sqrt{x-1}\sqrt{x+1}$ Jul 24, 2016 at 2:23

hint: $x-1 = (\sqrt{x}-1)(\sqrt{x}+1)$
$\frac{\frac{1}{\sqrt{x}}-1}{x-1}\cdot\frac{\sqrt{x}}{\sqrt{x}}=\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}=\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}=-\frac{1}{\sqrt{x}(\sqrt{x}+1)}$
• Man, it was that first step (multiplying top and bottom by $\sqrt{x}$ that I missed. ...getting back into math and teaching myself precalc/calc again. One step at a time! Jul 24, 2016 at 3:06