Chapter 9 - Roots and Radicals - 9.4 - Products and Quotients Involving Radicals - Problem Set 9.4 - Page 418: 64

$\dfrac{7\sqrt{x}-28}{x-16}$

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $ \dfrac{7}{\sqrt{x}+4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{7}{\sqrt{x}+4}\cdot\dfrac{\sqrt{x}-4}{\sqrt{x}-4} \\\\= \dfrac{7\sqrt{x}-28}{(\sqrt{x})^2-(4)^2} \\\\= \dfrac{7\sqrt{x}-28}{x-16} .\end{array}