Question
Answer
$$\dfrac{4}{2\sqrt{2}-4\sqrt{5}}=-\dfrac{\sqrt{2}+2\sqrt{5}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4}{2\sqrt{2}-4\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{2\sqrt{2}-4\sqrt{5}}\frac{2\sqrt{2}+4\sqrt{5}}{2\sqrt{2}+4\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{8\sqrt{2}+16\sqrt{5}}{8+8\sqrt{10}-8\sqrt{10}-80} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{8\sqrt{2}+16\sqrt{5}}{-72} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{2}+2\sqrt{5}}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{2}+2\sqrt{5}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2} + 4 \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 2 \sqrt{2} + 4 \sqrt{5}\right) = \color{blue}{4} \cdot 2 \sqrt{2}+\color{blue}{4} \cdot 4 \sqrt{5} = \\ = 8 \sqrt{2} + 16 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2}- 4 \sqrt{5}\right) } \cdot \left( 2 \sqrt{2} + 4 \sqrt{5}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot 4 \sqrt{5}\color{blue}{- 4 \sqrt{5}} \cdot 2 \sqrt{2}\color{blue}{- 4 \sqrt{5}} \cdot 4 \sqrt{5} = \\ = 8 + 8 \sqrt{10}- 8 \sqrt{10}-80 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 8. |
| ⑤ | Place a negative sign in front of a fraction. |
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