Answer
$2\sqrt{3}+\sqrt{10}-3\sqrt{2}-\sqrt{15}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}}
,$ 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{2}-\sqrt{3}}{\sqrt{6}-\sqrt{5}} \cdot\dfrac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}
\\\\=
\dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})}
.\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{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{(\sqrt{6})^2-(\sqrt{5})^2}
\\\\=
\dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{6-5}
\\\\=
\dfrac{(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})}{1}
\\\\=
(\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{5})
.\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{2}(\sqrt{6})+\sqrt{2}(\sqrt{5})-\sqrt{3}(\sqrt{6})-\sqrt{3}(\sqrt{5})
.\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{2(6)}+\sqrt{2(5)}-\sqrt{3(6)}-\sqrt{3(5)}
\\\\=
\sqrt{12}+\sqrt{10}-\sqrt{18}-\sqrt{15}
.\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{4\cdot3}+\sqrt{10}-\sqrt{9\cdot2}-\sqrt{15}
\\\\=
\sqrt{(2)^2\cdot3}+\sqrt{10}-\sqrt{(3)^2\cdot2}-\sqrt{15}
\\\\=
2\sqrt{3}+\sqrt{10}-3\sqrt{2}-\sqrt{15}
.\end{array}