Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4 - Page 418: 65

$\dfrac{x-3\sqrt{x}}{x-9}$

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $ \dfrac{\sqrt{x}}{\sqrt{x}+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-3} \\\\= \dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x})^2-(3)^2} \\\\= \dfrac{x-3\sqrt{x}}{x-9} .\end{array}