Question
Answer
$$\sqrt{6}\cdot(6+\sqrt{21})=6\sqrt{6}+3\sqrt{14}$$
Explanation
| $$ \begin{aligned}\sqrt{6}\cdot(6+\sqrt{21})& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}6\sqrt{6}+3\sqrt{14}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \sqrt{6} } \cdot \left( 6 + \sqrt{21}\right) = \color{blue}{ \sqrt{6}} \cdot6+\color{blue}{ \sqrt{6}} \cdot \sqrt{21} = \\ = 6 \sqrt{6} + 3 \sqrt{14} $$ |
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