Answer
$\dfrac{7\sqrt{ 2}}{6}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given radical expression, $
\sqrt{\dfrac{8}{9}}+\sqrt{\dfrac{18}{36}}
,$ find a factor of the radicand that is a perfect power of the index. Then, extract the root of that factor. Finally, combine the like radicals.
$\bf{\text{Solution Details:}}$
Rewriting the radicand with a factor that is a perfect power of the index, the given expression is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{\dfrac{4}{9}\cdot2}+\sqrt{\dfrac{9}{36}\cdot2}
\\\\=
\sqrt{\dfrac{4}{9}\cdot2}+\sqrt{\dfrac{1}{4}\cdot2}
\\\\=
\sqrt{\left( \dfrac{2}{3} \right)^2\cdot2}+\sqrt{ \left( \dfrac{1}{2} \right)^2\cdot2}
.\end{array}
Extracting the root of the factor that is a perfect power of the index results to
\begin{array}{l}\require{cancel}
\dfrac{2}{3}\sqrt{ 2}+\dfrac{1}{2}\sqrt{ 2}
\\\\=
\dfrac{2\sqrt{ 2}}{3}+\dfrac{\sqrt{ 2}}{2}
.\end{array}
To simplify the like radicals above, change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $
3
$ and $
2
$ is $
6
$ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to
\begin{array}{l}\require{cancel}
\dfrac{2\sqrt{ 2}}{3}\cdot\dfrac{2}{2}+\dfrac{\sqrt{ 2}}{2}\cdot\dfrac{3}{3}
\\\\=
\dfrac{4\sqrt{ 2}}{6}+\dfrac{3\sqrt{ 2}}{6}
\\\\=
\dfrac{4\sqrt{ 2}+3\sqrt{ 2}}{6}
\\\\=
\dfrac{7\sqrt{ 2}}{6}
.\end{array}
