Question
Answer
$$\dfrac{5+\sqrt{27}}{6-\sqrt{3}}=\dfrac{39+23\sqrt{3}}{33}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5+\sqrt{27}}{6-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5+\sqrt{27}}{6-\sqrt{3}}\frac{6+\sqrt{3}}{6+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{30+5\sqrt{3}+18\sqrt{3}+9}{36+6\sqrt{3}-6\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{39+23\sqrt{3}}{33}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5 + \sqrt{27}\right) } \cdot \left( 6 + \sqrt{3}\right) = \color{blue}{5} \cdot6+\color{blue}{5} \cdot \sqrt{3}+\color{blue}{ \sqrt{27}} \cdot6+\color{blue}{ \sqrt{27}} \cdot \sqrt{3} = \\ = 30 + 5 \sqrt{3} + 18 \sqrt{3} + 9 $$ Simplify denominator. $$ \color{blue}{ \left( 6- \sqrt{3}\right) } \cdot \left( 6 + \sqrt{3}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot6\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 36 + 6 \sqrt{3}- 6 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
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