Question
Answer
$$\dfrac{\sqrt{6}\cdot(\sqrt{2}-\sqrt{3})}{1}=2\sqrt{3}-3\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{6}\cdot(\sqrt{2}-\sqrt{3})}{1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{3}-3\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}2\sqrt{3}-3\sqrt{2}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \sqrt{6} } \cdot \left( \sqrt{2}- \sqrt{3}\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{2}+\color{blue}{ \sqrt{6}} \cdot- \sqrt{3} = \\ = 2 \sqrt{3}- 3 \sqrt{2} $$ |
| ② | Remove 1 from denominator. |
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