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