Answer
$$\frac{3+\sqrt{2}}{\sqrt{6}+3}=\frac{-3\sqrt{6}+9-2\sqrt{3}+3\sqrt{2}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{2}}{\sqrt{6}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{2}}{\sqrt{6}+3}\frac{\sqrt{6}-3}{\sqrt{6}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{6}-9+2\sqrt{3}-3\sqrt{2}}{6-3\sqrt{6}+3\sqrt{6}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{6}-9+2\sqrt{3}-3\sqrt{2}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-3\sqrt{6}+9-2\sqrt{3}+3\sqrt{2}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{2}\right) } \cdot \left( \sqrt{6}-3\right) = \color{blue}{3} \cdot \sqrt{6}+\color{blue}{3} \cdot-3+\color{blue}{ \sqrt{2}} \cdot \sqrt{6}+\color{blue}{ \sqrt{2}} \cdot-3 = \\ = 3 \sqrt{6}-9 + 2 \sqrt{3}- 3 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6} + 3\right) } \cdot \left( \sqrt{6}-3\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot-3+\color{blue}{3} \cdot \sqrt{6}+\color{blue}{3} \cdot-3 = \\ = 6- 3 \sqrt{6} + 3 \sqrt{6}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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