Answer
$$\frac{\sqrt{27}}{6}=\frac{\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{27}}{6}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{ \sqrt{ 9 \cdot 3 } }{ 6 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{ \sqrt{ 9 } \cdot \sqrt{ 3 } }{ 6 } \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{3}}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}} \frac{ 3 \cdot \sqrt{ 3 } : \color{orangered}{ 3 }}{ 6 : \color{orangered}{ 3 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{\sqrt{3}}{2}\end{aligned} $$ | |
| ① | Factor out the largest perfect square of 27. ( in this example we factored out $ 9 $ ) |
| ② | Rewrite $ \sqrt{ 9 \cdot 3 } $ as the product of two radicals. |
| ③ | The square root of $ 9 $ is $ 3 $. |
| ④ | Divide numerator and denominator by $ \color{orangered}{ 3 } $. |
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