DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$

Resolução:
$\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,=$ $\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,\centerdot \,\dfrac{\sqrt{a\,}\,+\,\sqrt{b\,}}{\;\sqrt{a\,}\,+\,\sqrt{b\,}\;}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(\sqrt{a\,})^2\,-\,(\sqrt{b\,})^2}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(a\,-\,b)}\,=\,$$\,\sqrt{a\,}\,+\,\sqrt{b\,} $