Question
Answer
$$\dfrac{(4\sqrt{3}-2\sqrt{2})(3\sqrt{2}+4\sqrt{3})}{1}=4\sqrt{6}+36$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(4\sqrt{3}-2\sqrt{2})(3\sqrt{2}+4\sqrt{3})}{1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{12\sqrt{6}+48-12-8\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4\sqrt{6}+36\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 4 \sqrt{3}- 2 \sqrt{2}\right) } \cdot \left( 3 \sqrt{2} + 4 \sqrt{3}\right) = \color{blue}{ 4 \sqrt{3}} \cdot 3 \sqrt{2}+\color{blue}{ 4 \sqrt{3}} \cdot 4 \sqrt{3}\color{blue}{- 2 \sqrt{2}} \cdot 3 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot 4 \sqrt{3} = \\ = 12 \sqrt{6} + 48-12- 8 \sqrt{6} $$ |
| ② | Remove 1 from denominator. |
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