Question
Answer
$$\dfrac{1}{\sqrt{8}+1}=\dfrac{2\sqrt{2}-1}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{8}+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{8}+1}\frac{\sqrt{8}-1}{\sqrt{8}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{2}-1}{8-2\sqrt{2}+2\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{2}-1}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{8}-1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{8}-1\right) = \color{blue}{1} \cdot \sqrt{8}+\color{blue}{1} \cdot-1 = \\ = 2 \sqrt{2}-1 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{8} + 1\right) } \cdot \left( \sqrt{8}-1\right) = \color{blue}{ \sqrt{8}} \cdot \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot-1+\color{blue}{1} \cdot \sqrt{8}+\color{blue}{1} \cdot-1 = \\ = 8- 2 \sqrt{2} + 2 \sqrt{2}-1 $$ |
| ③ | Simplify numerator and denominator |
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