<s>[INST] Question on an algebraic expression Why this is true?
$$
\frac{\frac{1}{\sqrt{x}}-1}{x-1} = -\frac{1}{\sqrt{x}+x}
$$
 [/INST] When you have complicated fractions, it is always a good idea to try making common denominators where applicable. The numerator is a bit complicated in this case, so let's work with that:
$$\frac{1}{\sqrt{x}} - 1 = \frac{1}{\sqrt{x}} - \frac{\sqrt{x}}{\sqrt{x}} = \frac{1-\sqrt{x}}{\sqrt{x}}.$$
Now let's replace $\frac{1}{\sqrt{x}} - 1$ with $\frac{1-\sqrt{x}}{\sqrt{x}}$ in your expression.
$$\frac{\frac{1}{\sqrt{x}}-1}{x-1} = \frac{\frac{1-\sqrt{x}}{\sqrt{x}}}{x-1}.$$
Moving the $\sqrt{x}$ to the denominator, we have
$$\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}.$$
In the numerator on the right hand side of your expression, you have $1$ but we have $1-\sqrt{x}$. What we should do now is rationalize the numerator by multiplying by $\frac{1+\sqrt{x}}{1+\sqrt{x}}$. Note that this does not change our expression any since it is really just a clever form of $1$. Doing so, we have
$$\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}\frac{1+\sqrt{x}}{1+\sqrt{x}} = \frac{(1-\sqrt{x})(1+\sqrt{x})}{\sqrt{x}(x-1)(1+\sqrt{x})} = \frac{1-x}{\sqrt{x}(x-1)(1+\sqrt{x})}.$$
Cancelling the common term of $1-x$ from the numerator and denominator gives us
$$-\frac{1}{\sqrt{x}(1+\sqrt{x})} = -\frac{1}{\sqrt{x}+x}$$
as desired.
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