Simple algebra. How can you factor out a common factor if it is $\frac{1}{\text{factor}}$ in one of the cases? I'm sure this is simple.  I want to pull out a factor as follows...
I have the expression
$$\frac{a(\sqrt{x}) - (b + c)(\frac{1}{\sqrt{x}})c}{x}.$$
It would be useful for me to pull out the $\sqrt{x}$ from the numerator and try to simplify to remove the denominator, but how can I pull out the $\sqrt{x}$ from the right-most statement $\frac{1}{\sqrt{x}}$.
Thanks for your help!

$$ \frac{a(\sqrt{x}) - (b + c)({\frac{\sqrt{x}}{x}})c}{x}\;\tag{1}$$
$$=\frac{a(\color{red}{\sqrt{x}}) - (b + c)({\frac{\color{red}{\sqrt{x}}}{x}})c}{\color{red}{\sqrt x}\sqrt x}\;\tag{2}$$
$$=\frac{(\color{red}{\sqrt x})[a - (b + c)({\frac{1}{x}})c]}{\color{red}{\sqrt x}\sqrt x}\;\tag{3}$$
$$=\frac{[a - (b + c)({\frac{1}{x}})c]}{\sqrt{x}}$$
And finally:
$$=\frac{[ax - (b + c)c]}{x\sqrt{x}}$$
Hope it helps.
$(1)$: Rewriting ${1\over\sqrt x}$ as ${\sqrt x \over x}$
$(2)$: Rewriting $x$ as $\sqrt x \sqrt x$, now $\sqrt x$ is the common factor in both numerator and denominator
$(3)$: Pulling the common factor ($\sqrt x$) out in the numerator
