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