Answer
$-\dfrac{89+42\sqrt{3}}{239}$
Work Step by Step
Extracting the root of the factor that is a perfect power of the index, the given expression, $
\dfrac{4+\sqrt{27}}{2-3\sqrt{27}}
,$ is equivalent to
\begin{align*}
&
\dfrac{4+\sqrt{9\cdot3}}{2-3\sqrt{9\cdot3}}
\\\\&=
\dfrac{4+\sqrt{3^2\cdot3}}{2-3\sqrt{3^2\cdot3}}
\\\\&=
\dfrac{4+3\sqrt{3}}{2-3(3)\sqrt{3}}
\\\\&=
\dfrac{4+3\sqrt{3}}{2-9\sqrt{3}}
.\end{align*}
Multiplying the numerator and the denominator by the conjugate of the denominator, the expression above is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{4+3\sqrt{3}}{2-9\sqrt{3}}\cdot\dfrac{2+9\sqrt{3}}{2+9\sqrt{3}}
\\\\&=
\dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{(2)^2-(9\sqrt{3})^2}
&\left( \text{use }(a+b)(a-b)=a^2-b^2 \right)
\\\\&=
\dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{4-81(3)}
\\\\&=
\dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{4-243}
\\\\&=
\dfrac{(4+3\sqrt{3})(2+9\sqrt{3})}{-239}
\\\\&=
\dfrac{4(2)+4(9\sqrt{3})+3\sqrt{3}(2)+3\sqrt{3}(9\sqrt{3})}{-239} &\left( \text{use FOIL} \right)
\\\\&=
\dfrac{8+36\sqrt{3}+6\sqrt{3}+27(3)}{-239}
\\\\&=
\dfrac{8+36\sqrt{3}+6\sqrt{3}+81}{-239}
\\\\&=
\dfrac{89+42\sqrt{3}}{-239}
\\\\&=
-\dfrac{89+42\sqrt{3}}{239}
.\end{align*}
Hence, the rationalized-denominator form of the given expression is $
-\dfrac{89+42\sqrt{3}}{239}
$.
