Question
Answer
$$\dfrac{5}{4\sqrt{2}-1}=\dfrac{20\sqrt{2}+5}{31}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5}{4\sqrt{2}-1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{4\sqrt{2}-1}\frac{4\sqrt{2}+1}{4\sqrt{2}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{20\sqrt{2}+5}{32+4\sqrt{2}-4\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{20\sqrt{2}+5}{31}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 \sqrt{2} + 1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 5 } \cdot \left( 4 \sqrt{2} + 1\right) = \color{blue}{5} \cdot 4 \sqrt{2}+\color{blue}{5} \cdot1 = \\ = 20 \sqrt{2} + 5 $$ Simplify denominator. $$ \color{blue}{ \left( 4 \sqrt{2}-1\right) } \cdot \left( 4 \sqrt{2} + 1\right) = \color{blue}{ 4 \sqrt{2}} \cdot 4 \sqrt{2}+\color{blue}{ 4 \sqrt{2}} \cdot1\color{blue}{-1} \cdot 4 \sqrt{2}\color{blue}{-1} \cdot1 = \\ = 32 + 4 \sqrt{2}- 4 \sqrt{2}-1 $$ |
| ③ | Simplify numerator and denominator |
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