College Algebra (11th Edition)

$\dfrac{-7\sqrt{3}}{36}$
$\bf{\text{Solution Outline:}}$ Simplify each term in the given expression, $\dfrac{2}{\sqrt{12}}-\dfrac{1}{\sqrt{27}}-\dfrac{5}{\sqrt{48}} .$ Then make the terms similar (same denominator) to combine the numerators. $\bf{\text{Solution Details:}}$ Extracting the perfect square factors of the radicals results to \begin{array}{l}\require{cancel} \dfrac{2}{\sqrt{4\cdot3}}-\dfrac{1}{\sqrt{9\cdot3}}-\dfrac{5}{\sqrt{16\cdot3}} \\\\= \dfrac{2}{\sqrt{(2)^2\cdot3}}-\dfrac{1}{\sqrt{(3)^2\cdot3}}-\dfrac{5}{\sqrt{(4)^2\cdot3}} \\\\= \dfrac{2}{2\sqrt{3}}-\dfrac{1}{3\sqrt{3}}-\dfrac{5}{4\sqrt{3}} .\end{array} Rationalizing the denominators results to \begin{array}{l}\require{cancel} \dfrac{2}{2\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{1}{3\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}-\dfrac{5}{4\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\\\= \dfrac{2\sqrt{3}}{2(\sqrt{3})^2}-\dfrac{\sqrt{3}}{3(\sqrt{3})^2}-\dfrac{5\sqrt{3}}{4(\sqrt{3})^2} \\\\= \dfrac{\cancel2\sqrt{3}}{\cancel2(3)}-\dfrac{\sqrt{3}}{3(3)}-\dfrac{5\sqrt{3}}{4(3)} \\\\= \dfrac{\sqrt{3}}{3}-\dfrac{\sqrt{3}}{9}-\dfrac{5\sqrt{3}}{12} .\end{array} To simplify the expression above, find the $LCD$ of the denominators. The $LCD$ is $36$ since it is the lowest number that can be exactly divided by the denominators $3,9, \text{ and } 12 .$ Multiplying each term by an expression equal to $1$ so that its denominator becomes the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{\sqrt{3}}{3}\cdot\dfrac{12}{12}-\dfrac{\sqrt{3}}{9}\cdot\dfrac{4}{4}-\dfrac{5\sqrt{3}}{12}\cdot\dfrac{3}{3} \\\\= \dfrac{12\sqrt{3}}{36}-\dfrac{4\sqrt{3}}{36}-\dfrac{15\sqrt{3}}{36} .\end{array} To combine similar terms, add/subtract the numerators and copy the similar denominator. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{12\sqrt{3}-4\sqrt{3}-15\sqrt{3}}{36} \\\\= \dfrac{-7\sqrt{3}}{36} .\end{array}