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This question already has an answer here:

I know this looks like a dumb question, but how do I simplify this? Does it uses some square root property or factorization? The wolfram alpha has no step-by-step solution for this, so it may use some crazy technique, I guess.

$$\frac{\sqrt{4+h}-2}{h} = \frac{1}{\sqrt{4+h}-2}$$

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marked as duplicate by user63181, user127096, user61527, Dan, Hans Engler Mar 30 '14 at 0:27

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    $\begingroup$ It is simple already, but very useful is to multiply by $\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$. We get $\frac{1}{\sqrt{4+h}+2}$. $\endgroup$ – André Nicolas Mar 29 '14 at 21:37
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$$\dfrac{\sqrt{4+h}-2}{h}$$ $$=\dfrac{\sqrt{4+h}-2}{h}\times\dfrac{\sqrt{4+h}+2}{\sqrt{4+h}+2}$$ $$=\dfrac{(4+h)-4}{h(\sqrt{4+h}+2)}$$ $$=\dfrac{h}{h(\sqrt{4+h}+2)}$$ $$=\dfrac{1}{\sqrt{4+h}+2}$$

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Hint: Multiply with $$\sqrt{4+h}+2$$ both the numerator and the denominator. The RHS shoud be $$\frac{1}{\sqrt{4+h}+2}$$ and not with $-2$ as you have it.

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$$\frac{\sqrt{4+h}-2}{h}=\frac{(\sqrt{4+h}-2)\cdot(\sqrt{4+h}+2)}{h\cdot(\sqrt{4+h}+2)}=\frac{(\sqrt{4+h})^2-4}{h(\sqrt{4+h}+2)} = \frac{1}{\sqrt{4+h}+2}$$ Obviously we suppose $h\ne0$.

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