Answer
$$\frac{(4+2\sqrt{3})\cdot(4-2\sqrt{3})}{\sqrt{11}}=\frac{4\sqrt{11}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(4+2\sqrt{3})\cdot(4-2\sqrt{3})}{\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{16-8\sqrt{3}+8\sqrt{3}-12}{\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4}{\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}} \frac{ 4 }{\sqrt{ 11 }} \times \frac{ \color{orangered}{\sqrt{ 11 }} }{ \color{orangered}{\sqrt{ 11 }}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4\sqrt{11}}{11}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 4 + 2 \sqrt{3}\right) } \cdot \left( 4- 2 \sqrt{3}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot4+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 16- 8 \sqrt{3} + 8 \sqrt{3}-12 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply both top and bottom by $ \color{orangered}{ \sqrt{ 11 }}$. |
| ④ | In denominator we have $ \sqrt{ 11 } \cdot \sqrt{ 11 } = 11 $. |
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