Chapter 9 - Roots and Radicals - Chapters 1-9 Cumulative Review Problem Set - Page 432: 11

$-\dfrac{15}{4x}$

Using the $LCD= 4x $, the given expression, $ \dfrac{3}{4x}+\dfrac{5}{2x}-\dfrac{7}{x} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{1(3)+2(5)-4(7)}{4x} \\\\= \dfrac{3+10-28}{4x} \\\\= \dfrac{-15}{4x} \\\\= -\dfrac{15}{4x} .\end{array}