Difference of the square roots of consecutive integers is equal to the reciprocal of the sum of their square roots

Question

The difference of the square roots of consecutive integers is equal to the reciprocal of the sum of the square roots of the same integers, like shown below.

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

However, I'm interested in looking at the proof for the above but I am unable to find any online or in a book. Can anyone show me?

Thank you very much

Answer 1

$\sqrt{x+1}-\sqrt{x}= \frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x}) }{\sqrt{x+1}+\sqrt{x}}.$

Can you proceed ?

Answer 2

$$(\sqrt{x+1}-\sqrt{x})( {\sqrt{x+1}+\sqrt{x}})=(x+1)-x=1$$