Answer
$$\frac{3}{2-\sqrt{6}}=-\frac{6+3\sqrt{6}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{2-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{2-\sqrt{6}}\frac{2+\sqrt{6}}{2+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6+3\sqrt{6}}{4+2\sqrt{6}-2\sqrt{6}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6+3\sqrt{6}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{6+3\sqrt{6}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( 2 + \sqrt{6}\right) = \color{blue}{3} \cdot2+\color{blue}{3} \cdot \sqrt{6} = \\ = 6 + 3 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{6}\right) } \cdot \left( 2 + \sqrt{6}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot2\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 4 + 2 \sqrt{6}- 2 \sqrt{6}-6 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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