Answer
$$\frac{\sqrt{27}+\sqrt{3}}{\sqrt{2}}=2\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{27}+\sqrt{3}}{\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{27}+\sqrt{3}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{6}+\sqrt{6}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4\sqrt{6}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ }2\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{27} + \sqrt{3}\right) } \cdot \sqrt{2} = \color{blue}{ \sqrt{27}} \cdot \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot \sqrt{2} = \\ = 3 \sqrt{6} + \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \sqrt{2} } \cdot \sqrt{2} = 2 $$ |
| ③ | Simplify numerator and denominator |
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