Answer
$$\frac{(6+\sqrt{3})\cdot(6-\sqrt{3})}{\sqrt{3}\cdot3}=\frac{11\sqrt{3}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(6+\sqrt{3})\cdot(6-\sqrt{3})}{\sqrt{3}\cdot3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{36-6\sqrt{3}+6\sqrt{3}-3}{\sqrt{3}\cdot3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{33}{3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{33}{3\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{33\sqrt{3}}{9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{ 33 \sqrt{ 3 } : \color{blue}{ 3 } } { 9 : \color{blue}{ 3 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{11\sqrt{3}}{3}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 6 + \sqrt{3}\right) } \cdot \left( 6- \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot6+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 36- 6 \sqrt{3} + 6 \sqrt{3}-3 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 33 } \cdot \sqrt{3} = 33 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ 3 \sqrt{3} } \cdot \sqrt{3} = 9 $$ |
| ⑤ | Divide numerator and denominator by $ \color{blue}{ 3 } $. |
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