Question
Answer
$$\sqrt{3}\cdot(\sqrt{6}-\sqrt{12})=3\sqrt{2}-6$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\sqrt{3}\cdot(\sqrt{6}-\sqrt{12})& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\sqrt{3}\cdot(\sqrt{6}-2\sqrt{3}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}3\sqrt{2}-6\end{aligned} $$ | |
| ① | $$ - \sqrt{12} =
- \sqrt{ 2 ^2 \cdot 3 } =
- \sqrt{ 2 ^2 } \, \sqrt{ 3 } =
- 2 \sqrt{ 3 }$$ |
| ② | $$ \color{blue}{ \sqrt{3} } \cdot \left( \sqrt{6}- 2 \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{6}+\color{blue}{ \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 3 \sqrt{2}-6 $$ |
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