Answer
$$\frac{27}{\sqrt{12}}=\frac{9\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{27}{\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}} \frac{ 27 }{\sqrt{ 12 }} \times \frac{ \color{orangered}{\sqrt{ 12 }} }{ \color{orangered}{\sqrt{ 12 }}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{27\sqrt{12}}{12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}} \frac{ 27 \sqrt{ 4 \cdot 3 }}{ 12 } \xlongequal{ } \\[1 em] & \xlongequal{ } \frac{ 27 \cdot 2 \sqrt{ 3 } }{ 12 } \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{54\sqrt{3}}{12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}} \frac{ 54 \sqrt{ 3 } : \color{blue}{ 6 } }{ 12 : \color{blue}{ 6 } } \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{9\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Multiply both top and bottom by $ \color{orangered}{ \sqrt{ 12 }}$. |
| ② | In denominator we have $ \sqrt{ 12 } \cdot \sqrt{ 12 } = 12 $. |
| ③ | Simplify $ \sqrt{ 12 } $. |
| ④ | Divide both the top and bottom numbers by $ \color{blue}{ 6 }$. |
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