#### Answer

$\sqrt{15}+\sqrt{10}+3\sqrt{2}+2\sqrt{3}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{3}-\sqrt{2}}
,$ multiply the numerator and the denominator by the conjugate of the denominator. Then use special products to simplify the result.
$\bf{\text{Solution Details:}}$
Multiplying the numerator and the denominator of the given expression by the conjugate of the denominator, the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{3}-\sqrt{2}} \cdot\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}
\\\\=
\dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}
.\end{array}
Using the product of the sum and difference of like terms which is given by $(a+b)(a-b)=a^2-b^2,$ the expression above is equivalent
\begin{array}{l}\require{cancel}
\dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}
\\\\=
\dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{3-2}
\\\\=
\dfrac{(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})}{1}
\\\\=
(\sqrt{5}+\sqrt{6})(\sqrt{3}+\sqrt{2})
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to\begin{array}{l}\require{cancel}
\sqrt{5}(\sqrt{3})+\sqrt{5}(\sqrt{2})+\sqrt{6}(\sqrt{3})+\sqrt{6}(\sqrt{2})
.\end{array}
Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{5(3)}+\sqrt{5(2)}+\sqrt{6(3)}+\sqrt{6(2)}
\\\\=
\sqrt{15}+\sqrt{10}+\sqrt{18}+\sqrt{12}
.\end{array}
Writing the radicand as an expression that contains a factor that is a perfect power of the index and extracting the root of that factor result to
\begin{array}{l}\require{cancel}
\sqrt{15}+\sqrt{10}+\sqrt{9\cdot2}+\sqrt{4\cdot3}
\\\\=
\sqrt{15}+\sqrt{10}+\sqrt{(3)^2\cdot2}+\sqrt{(2)^2\cdot3}
\\\\=
\sqrt{15}+\sqrt{10}+3\sqrt{2}+2\sqrt{3}
.\end{array}