Answer
$\dfrac{9+3\sqrt2}{7}$
Work Step by Step
Simplify each radical to obtain:
\begin{align*}
&=\dfrac{3+\sqrt{9(2)}}{1+\sqrt{4(2)}}\\\\
&=\dfrac{3+\sqrt{3^2(2)}}{1+\sqrt{2^2(2)}}\\\\
&=\dfrac{3+3\sqrt2}{1+2\sqrt2}
\end{align*}
Rationalize the denominator by multiplying $1-2\sqrt2$ to both the numerator and denominator to obtain:
\begin{align*}
&=\dfrac{3+3\sqrt2}{1+2\sqrt2}\cdot \dfrac{1-2\sqrt2}{1-2\sqrt2}\\\\
\end{align*}
Recall:
$(a+b)(a-b) = a^2-b^2$
Use the rule above with $a=1$ and $b=2\sqrt{2}$ to obtain:
\begin{align*}
\dfrac{3+3\sqrt2}{1+2\sqrt2}\cdot \dfrac{1-2\sqrt2}{1-2\sqrt2}&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{1^2-(2\sqrt2)^2}\\\\
&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{1^2-(2\sqrt2)^2}\\\\
&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{1-2^2(\sqrt2)^2)}\\\\
&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{1-4(2)}\\\\
&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{1-8}\\\\
&=\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{-7}\\\\
\end{align*}
Recall:
$(a+b)(c+d)=a(c+d)+b(c=+)$
Use the rule above to simplify the numerator and obtain:
\begin{align*}
\dfrac{(3+3\sqrt2)(1-2\sqrt2)}{-7}&=\dfrac{3(1-2\sqrt2)+3\sqrt2(1-2\sqrt2)}{-7}\\\\
&=\dfrac{3(1)-3(2\sqrt2)+3\sqrt2(1)-3\sqrt2(2\sqrt2)}{-7}\\\\
&=\dfrac{3-6\sqrt2+3\sqrt2-6\sqrt4}{-7}\\\\
&=\dfrac{3-6\sqrt2+3\sqrt2-6(2)}{-7}\\\\
&=\dfrac{3-6\sqrt2+3\sqrt2-12}{-7}\\\\
\end{align*}
Combine like terms to obtain:
\begin{align*}
&=\dfrac{(3-12)+(-6\sqrt2+3\sqrt2)}{-7}\\\\
&=\dfrac{-9+(-6+3)(\sqrt2)}{-7}\\\\
&=\dfrac{-9+(-3)(\sqrt2)}{-7}\\\\
&=\dfrac{-9-3\sqrt2}{-7}\\\\
&=\dfrac{-(9+3\sqrt2)}{-7}\\\\
&=\dfrac{9+3\sqrt2}{7}
\end{align*}
