Answer
$$\frac{\sqrt{2}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}=\frac{3\sqrt{6}-2\sqrt{3}-9+3\sqrt{2}}{21}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}-\sqrt{3}}{3\sqrt{3}+\sqrt{6}}\frac{3\sqrt{3}-\sqrt{6}}{3\sqrt{3}-\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{6}-2\sqrt{3}-9+3\sqrt{2}}{27-9\sqrt{2}+9\sqrt{2}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{6}-2\sqrt{3}-9+3\sqrt{2}}{21}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3}- \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2}- \sqrt{3}\right) } \cdot \left( 3 \sqrt{3}- \sqrt{6}\right) = \color{blue}{ \sqrt{2}} \cdot 3 \sqrt{3}+\color{blue}{ \sqrt{2}} \cdot- \sqrt{6}\color{blue}{- \sqrt{3}} \cdot 3 \sqrt{3}\color{blue}{- \sqrt{3}} \cdot- \sqrt{6} = \\ = 3 \sqrt{6}- 2 \sqrt{3}-9 + 3 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3} + \sqrt{6}\right) } \cdot \left( 3 \sqrt{3}- \sqrt{6}\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot- \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot 3 \sqrt{3}+\color{blue}{ \sqrt{6}} \cdot- \sqrt{6} = \\ = 27- 9 \sqrt{2} + 9 \sqrt{2}-6 $$ |
| ③ | Simplify numerator and denominator |
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