Answer
$\color{blue}{\dfrac{-7\sqrt3}{36}}$
Work Step by Step
Simplify the denominators to obtain:
$=\dfrac{2}{\sqrt{4(3)}} - \dfrac{1}{\sqrt{9(3)}} - \dfrac{5}{\sqrt{16(3)}}
\\=\dfrac{2}{2\sqrt3} - \dfrac{1}{3\sqrt{3}}- \dfrac{5}{4\sqrt{3}}
\\=\dfrac{1}{\sqrt3} - \dfrac{1}{3\sqrt{3}}- \dfrac{5}{4\sqrt{3}}$
Make the expressions similar by using their LCD of $12\sqrt{3}$ to obtain:
$=\dfrac{1(12)}{\sqrt3(12)}- \dfrac{1(4)}{3\sqrt3(4)}-\dfrac{5(3)}{4\sqrt{3}(3)}
\\=\dfrac{12}{12\sqrt3}-\dfrac{4}{12\sqrt3}-\dfrac{15}{12\sqrt3}$
Subtract the second and third numerators to the first and copy the denominator to obtain:
$=\dfrac{12-4-15}{12\sqrt3}
\\=\dfrac{-7}{12\sqrt3}$
Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain:
$=\dfrac{-7(\sqrt3)}{12\sqrt3(\sqrt3)}
\\=\dfrac{-7\sqrt3}{12(3)}
\\=\color{blue}{\dfrac{-7\sqrt3}{36}}$
