#### Answer

$\dfrac{a+2\sqrt{ab}+b}{a-b}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To rationalize the given radical expression, $
\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}
,$ 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{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}
\\\\=
\dfrac{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}
\\\\=
\dfrac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}
.\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{a}+\sqrt{b})^2}{(\sqrt{a})^2-(\sqrt{b})^2}
\\\\=
\dfrac{(\sqrt{a}+\sqrt{b})^2}{a-b}
.\end{array}
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{(\sqrt{a})^2+2(\sqrt{a})(\sqrt{b})+(\sqrt{b})^2}{a-b}
\\\\=
\dfrac{a+2(\sqrt{a})(\sqrt{b})+b}{a-b}
.\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}
\dfrac{a+2\sqrt{a(b)}+b}{a-b}
\\\\=
\dfrac{a+2\sqrt{ab}+b}{a-b}
.\end{array}