The following expression:
$$\frac{\sqrt{4+h}-2}{h}$$
should be simplified to:
$$\frac{1}{\sqrt{4+h}+2}$$
(even if I don't agree that this second is more simple than the first).
The problem is that I have no idea of the first step to simplify that.. any help?
On
It is really simple. Let us just do what is most intuitive, multiply numerator and denominator with what you want to have in denominator. You get: $$ \frac{(\sqrt{4+h} - 2)(\sqrt{4+h} + 2)}{h(\sqrt{4+h}+2)} $$ Then observe the numerator has a difference of squares. Multiply the numerator easily using that and then your left with $$\frac{h}{h(\sqrt{4+h}+2)}$$ Just assume $ h \neq 0 $ and get "rid" of it.
On
HINT $\rm\displaystyle\quad\quad g^2 = 4+h\ \ \Rightarrow\ \ \frac{g-2}h\ =\ \frac{g-2}{g^2-4}\ =\ \frac{1}{g+2}$
Usually the "simplification" is the opposite inference - known as rationalizing the denominator.
If you multiply both the top and the bottom by $\sqrt{4+h}+2$, you get $\frac{(\sqrt{4+h}-2)(\sqrt{4+h}+2)}{h(\sqrt{4+h}+2)}$, which simplifies to $\frac{h}{h(\sqrt{4+h}+2)}$. Then, divide both by $h$ (assuming $h\neq 0$), and you get $\frac{1}{\sqrt{4+h}+2}$.