Answer
$$\frac{2\sqrt{6}}{2\sqrt{27}-\sqrt{8}}=\frac{9\sqrt{2}+2\sqrt{3}}{25}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{6}}{2\sqrt{27}-\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{6}}{2\sqrt{27}-\sqrt{8}}\frac{2\sqrt{27}+\sqrt{8}}{2\sqrt{27}+\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36\sqrt{2}+8\sqrt{3}}{108+12\sqrt{6}-12\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{36\sqrt{2}+8\sqrt{3}}{100} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9\sqrt{2}+2\sqrt{3}}{25}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{27} + \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{6} } \cdot \left( 2 \sqrt{27} + \sqrt{8}\right) = \color{blue}{ 2 \sqrt{6}} \cdot 2 \sqrt{27}+\color{blue}{ 2 \sqrt{6}} \cdot \sqrt{8} = \\ = 36 \sqrt{2} + 8 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{27}- \sqrt{8}\right) } \cdot \left( 2 \sqrt{27} + \sqrt{8}\right) = \color{blue}{ 2 \sqrt{27}} \cdot 2 \sqrt{27}+\color{blue}{ 2 \sqrt{27}} \cdot \sqrt{8}\color{blue}{- \sqrt{8}} \cdot 2 \sqrt{27}\color{blue}{- \sqrt{8}} \cdot \sqrt{8} = \\ = 108 + 12 \sqrt{6}- 12 \sqrt{6}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
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